💡Okay, so I have this problem with two vectors, a and b. Vector a is (1, 2) and vector b is (2, -2). There are two parts to the problem. First, I need to find the dot product of vectors a and b. I remember that the dot product is calculated by multiplying the corresponding components of the vectors and then adding them together. So, for vectors a = (a1, a2) and b = (b1, b2), the dot product a · b is a1*b1 + a2*b2. Let me apply that to the given vectors. So, for vector a, the components are 1 and 2, and for vector b, the components are 2 and -2. Therefore, the dot product should be (1*2) + (2*(-2)). Calculating that, 1 times 2 is 2, and 2 times -2 is -4. Adding those together, 2 + (-4) equals -2. So, the dot product of a and b is -2. That seems straightforward.Now, moving on to the second part. It says that vector a + λ times vector b is perpendicular to vector a. I need to find the value of λ. Hmm, okay. I remember that if two vectors are perpendicular, their dot product is zero. So, if (a + λb) is perpendicular to a, then the dot product of (a + λb) and a should be zero.Let me write that down. So, (a + λb) · a = 0. I need to compute this dot product and solve for λ. First, let me find the expression for a + λb. Vector a is (1, 2), and vector b is (2, -2). So, multiplying vector b by λ gives (2λ, -2λ). Adding this to vector a, we get:a + λb = (1 + 2λ, 2 + (-2λ)) = (1 + 2λ, 2 - 2λ).Now, I need to take the dot product of this resulting vector with vector a. Vector a is (1, 2), so the dot product is:(1 + 2λ)*1 + (2 - 2λ)*2.Let me compute each part step by step. First, (1 + 2λ)*1 is just 1 + 2λ. Then, (2 - 2λ)*2 is 4 - 4λ. Adding these two results together: (1 + 2λ) + (4 - 4λ).Combining like terms, 1 + 4 is 5, and 2λ - 4λ is -2λ. So, the dot product simplifies to 5 - 2λ.Since the vectors are perpendicular, this dot product must equal zero. So, I set up the equation:5 - 2λ = 0.Solving for λ, I subtract 5 from both sides:-2λ = -5.Then, divide both sides by -2:λ = (-5)/(-2) = 5/2.So, λ is 5/2. Let me double-check my steps to make sure I didn't make any mistakes. First, calculating the dot product of a and b: 1*2 + 2*(-2) = 2 - 4 = -2. That seems correct.Then, for the second part, I expressed a + λb as (1 + 2λ, 2 - 2λ). Taking the dot product with a gives (1 + 2λ)*1 + (2 - 2λ)*2. Expanding that, I get 1 + 2λ + 4 - 4λ, which simplifies to 5 - 2λ. Setting that equal to zero gives λ = 5/2. Wait, let me verify the expansion again. (1 + 2λ)*1 is 1 + 2λ, and (2 - 2λ)*2 is 4 - 4λ. Adding them together: 1 + 2λ + 4 - 4λ. Combining like terms: 1 + 4 is 5, and 2λ - 4λ is -2λ. So, yes, 5 - 2λ = 0. Solving for λ, I get λ = 5/2. That seems consistent. I don't see any errors in my calculations. So, I think the value of λ is indeed 5/2.

💡First, I need to find the derivative of the given curve ( y = x^2 + ax + b ) to determine the slope of the tangent line at the point ( (0, b) ).Calculating the derivative:[y' = 2x + a]At ( x = 0 ), the slope of the tangent line is:[y' = 2(0) + a = a]Next, I'll rewrite the given tangent line equation ( x - y + 1 = 0 ) in slope-intercept form ( y = mx + c ) to identify its slope and y-intercept.Rewriting the equation:[x - y + 1 = 0 -y = -x - 1 y = x + 1]From this, the slope ( m ) of the tangent line is 1. Since the slope of the tangent line at ( (0, b) ) is also ( a ), I can set them equal:[a = 1]Now, substituting ( a = 1 ) back into the original curve equation:[y = x^2 + 1x + b]At the point ( (0, b) ), substituting ( x = 0 ) gives:[y = 0^2 + 1(0) + b = b]From the tangent line equation ( y = x + 1 ), when ( x = 0 ):[y = 0 + 1 = 1]Therefore, ( b = 1 ).So, the values of ( a ) and ( b ) are ( a = 1 ) and ( b = 1 ).

💡Okay, so I have this problem about a wheel with a circumference of 15 feet. The problem says that if the time for one complete rotation is reduced by 1/3 of a second, the speed of the wheel increases by 6 miles per hour. I need to find the original speed of the wheel in miles per hour. The options are 8, 9, 10, or 11. Hmm, let me try to figure this out step by step.First, I know that the circumference of the wheel is 15 feet. That means that for each full rotation, the wheel covers 15 feet. So, if I can find out how long it takes to make one rotation, I can figure out the speed.Let me denote the original time for one rotation as ( t ) seconds. Then, the original speed ( r ) in feet per second would be the distance covered divided by the time, so ( r = frac{15}{t} ) feet per second. But the problem asks for the speed in miles per hour, so I need to convert this.I remember that 1 mile is 5280 feet, and 1 hour is 3600 seconds. So, to convert feet per second to miles per hour, I can multiply by ( frac{3600}{5280} ). Let me write that down:Original speed in mph: ( r_{text{mph}} = frac{15}{t} times frac{3600}{5280} ).Simplifying that, ( frac{3600}{5280} ) is equal to ( frac{3600 ÷ 240}{5280 ÷ 240} = frac{15}{22} ). So, ( r_{text{mph}} = frac{15}{t} times frac{15}{22} = frac{225}{22t} ).Okay, so that's the original speed. Now, the problem says that if the time for one rotation is reduced by ( frac{1}{3} ) of a second, the speed increases by 6 mph. So, the new time is ( t - frac{1}{3} ) seconds, and the new speed is ( r_{text{mph}} + 6 ).Let me write the equation for the new speed:New speed in mph: ( r_{text{mph}} + 6 = frac{15}{t - frac{1}{3}} times frac{3600}{5280} ).Again, simplifying ( frac{3600}{5280} ) to ( frac{15}{22} ), so:( r_{text{mph}} + 6 = frac{15}{t - frac{1}{3}} times frac{15}{22} = frac{225}{22(t - frac{1}{3})} ).Now, I have two expressions for ( r_{text{mph}} ):1. ( r_{text{mph}} = frac{225}{22t} )2. ( r_{text{mph}} + 6 = frac{225}{22(t - frac{1}{3})} )I can substitute the first equation into the second one:( frac{225}{22t} + 6 = frac{225}{22(t - frac{1}{3})} ).Let me subtract ( frac{225}{22t} ) from both sides to isolate the 6:( 6 = frac{225}{22(t - frac{1}{3})} - frac{225}{22t} ).Factor out ( frac{225}{22} ):( 6 = frac{225}{22} left( frac{1}{t - frac{1}{3}} - frac{1}{t} right) ).Let me compute the expression inside the parentheses:( frac{1}{t - frac{1}{3}} - frac{1}{t} = frac{t - (t - frac{1}{3})}{t(t - frac{1}{3})} = frac{frac{1}{3}}{t(t - frac{1}{3})} ).So, substituting back:( 6 = frac{225}{22} times frac{frac{1}{3}}{t(t - frac{1}{3})} ).Simplify ( frac{225}{22} times frac{1}{3} = frac{225}{66} = frac{75}{22} ).So, ( 6 = frac{75}{22} times frac{1}{t(t - frac{1}{3})} ).Let me write that as:( 6 = frac{75}{22 t(t - frac{1}{3})} ).Multiply both sides by ( 22 t(t - frac{1}{3}) ):( 6 times 22 t(t - frac{1}{3}) = 75 ).Calculate ( 6 times 22 = 132 ):( 132 t(t - frac{1}{3}) = 75 ).Let me expand the left side:( 132 t^2 - 132 times frac{1}{3} t = 75 ).Simplify ( 132 times frac{1}{3} = 44 ):( 132 t^2 - 44 t = 75 ).Bring all terms to one side:( 132 t^2 - 44 t - 75 = 0 ).Now, I have a quadratic equation in terms of ( t ). Let me write it as:( 132 t^2 - 44 t - 75 = 0 ).To solve this quadratic equation, I can use the quadratic formula:( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ),where ( a = 132 ), ( b = -44 ), and ( c = -75 ).First, compute the discriminant:( b^2 - 4ac = (-44)^2 - 4 times 132 times (-75) ).Calculate ( (-44)^2 = 1936 ).Calculate ( 4 times 132 times (-75) = 4 times 132 times (-75) = 528 times (-75) = -39600 ).But since it's ( -4ac ), it becomes ( -4 times 132 times (-75) = +39600 ).So, discriminant is ( 1936 + 39600 = 41536 ).Now, take the square root of 41536. Let me see, 204 squared is 41616, which is a bit higher. 203 squared is 41209, which is a bit lower. Hmm, 204^2 = 41616, so sqrt(41536) is approximately 203.8, but maybe it's a whole number. Let me check:204^2 = 41616203^2 = 41209Difference between 41536 and 41209 is 327, which is not a perfect square. So, maybe it's not a whole number. Hmm, maybe I made a mistake in calculation.Wait, let me recalculate the discriminant:( b^2 = (-44)^2 = 1936 ).( 4ac = 4 times 132 times (-75) = 4 times 132 times (-75) ).Wait, 4 times 132 is 528, times (-75) is -39600. So, discriminant is ( 1936 - (-39600) = 1936 + 39600 = 41536 ). So that's correct.Hmm, sqrt(41536). Let me see, 204^2 is 41616, which is 80 more than 41536. So, sqrt(41536) is 204 - something. Maybe 203.8? Let me check 203.8^2:203.8^2 = (200 + 3.8)^2 = 200^2 + 2*200*3.8 + 3.8^2 = 40000 + 1520 + 14.44 = 41534.44. That's very close to 41536. So, approximately 203.8.So, sqrt(41536) ≈ 203.8.So, ( t = frac{-(-44) pm 203.8}{2 times 132} = frac{44 pm 203.8}{264} ).So, two solutions:1. ( t = frac{44 + 203.8}{264} = frac{247.8}{264} ≈ 0.938 ) seconds.2. ( t = frac{44 - 203.8}{264} = frac{-159.8}{264} ≈ -0.605 ) seconds.But time cannot be negative, so we discard the negative solution. So, ( t ≈ 0.938 ) seconds.Wait, but 0.938 seconds per rotation seems a bit fast for a wheel with a circumference of 15 feet. Let me check my calculations again.Wait, maybe I made a mistake in the quadratic equation setup.Let me go back.Original equation after simplifying:( 6 = frac{75}{22 t(t - frac{1}{3})} ).Multiply both sides by ( 22 t(t - frac{1}{3}) ):( 6 times 22 t(t - frac{1}{3}) = 75 ).Which is ( 132 t(t - frac{1}{3}) = 75 ).Expanding: ( 132 t^2 - 44 t = 75 ).So, ( 132 t^2 - 44 t - 75 = 0 ). That seems correct.Quadratic formula: ( t = frac{44 pm sqrt{(-44)^2 - 4 times 132 times (-75)}}{2 times 132} ).Which is ( t = frac{44 pm sqrt{1936 + 39600}}{264} = frac{44 pm sqrt{41536}}{264} ).As before, sqrt(41536) ≈ 203.8.So, ( t ≈ frac{44 + 203.8}{264} ≈ frac{247.8}{264} ≈ 0.938 ) seconds.Hmm, okay, maybe it's correct. So, the original time per rotation is approximately 0.938 seconds.Now, let's find the original speed in mph.Original speed ( r_{text{mph}} = frac{225}{22t} ).Plugging in ( t ≈ 0.938 ):( r_{text{mph}} ≈ frac{225}{22 times 0.938} ≈ frac{225}{20.636} ≈ 10.89 ) mph.Wait, but the options are 8, 9, 10, 11. 10.89 is close to 11, but let me check if I did everything correctly.Wait, maybe I should solve the quadratic equation more accurately.Let me compute sqrt(41536) more precisely.We know that 204^2 = 41616, which is 80 more than 41536.So, sqrt(41536) = 204 - ε, where ε is small.Let me compute (204 - ε)^2 = 41536.Expanding: 204^2 - 2*204*ε + ε^2 = 41536.We know 204^2 = 41616, so:41616 - 408ε + ε^2 = 41536.Subtract 41536:80 - 408ε + ε^2 = 0.Since ε is small, ε^2 is negligible, so approximately:80 - 408ε ≈ 0 ⇒ ε ≈ 80 / 408 ≈ 0.196.So, sqrt(41536) ≈ 204 - 0.196 ≈ 203.804.So, more accurately, sqrt(41536) ≈ 203.804.Thus, ( t = frac{44 + 203.804}{264} ≈ frac{247.804}{264} ≈ 0.9386 ) seconds.So, t ≈ 0.9386 seconds.Now, compute ( r_{text{mph}} = frac{225}{22t} ≈ frac{225}{22 times 0.9386} ≈ frac{225}{20.65} ≈ 10.89 ) mph.Hmm, so approximately 10.89 mph, which is close to 11 mph. But 10.89 is closer to 11, but the options include 10 and 11. Maybe I made a mistake in the setup.Wait, let me double-check the initial setup.Original speed: ( r = frac{15}{t} ) feet per second.Convert to mph: ( r_{text{mph}} = frac{15}{t} times frac{3600}{5280} = frac{15 times 3600}{5280 t} = frac{54000}{5280 t} = frac{54000 ÷ 240}{5280 ÷ 240 t} = frac{225}{22 t} ). That seems correct.New time: ( t - frac{1}{3} ) seconds.New speed: ( r + 6 = frac{15}{t - frac{1}{3}} times frac{3600}{5280} = frac{225}{22(t - frac{1}{3})} ).So, equation: ( frac{225}{22 t} + 6 = frac{225}{22(t - frac{1}{3})} ).Yes, that seems correct.Wait, maybe I should solve the quadratic equation exactly.We have ( 132 t^2 - 44 t - 75 = 0 ).Let me try to factor this equation.Looking for factors of ( 132 times (-75) = -9900 ) that add up to -44.Hmm, 9900 is a large number. Let me see, 9900 divided by 100 is 99, so maybe 99 and 100? Wait, 99 and 100 add up to 199, which is not helpful.Alternatively, maybe 132 t^2 - 44 t - 75 can be factored.Let me try to factor:Looking for two numbers that multiply to ( 132 times (-75) = -9900 ) and add to -44.Let me see, 9900 divided by 100 is 99, so maybe 99 and 100? Wait, 99 and 100 add up to 199, which is not helpful.Alternatively, 9900 = 99 * 100, but 99 and 100 don't add up to 44.Wait, maybe 9900 = 90 * 110, but 90 and 110 add up to 200.Hmm, this might not be factorable easily. Maybe I should use the quadratic formula.So, as before, ( t = frac{44 pm sqrt{41536}}{264} ).But sqrt(41536) is approximately 203.804, so:( t ≈ frac{44 + 203.804}{264} ≈ 0.9386 ) seconds.Alternatively, maybe I can express sqrt(41536) as 203.804, so:( t ≈ 0.9386 ) seconds.So, original speed ( r_{text{mph}} ≈ 10.89 ) mph.But the options are 8, 9, 10, 11. So, 10.89 is closer to 11, but maybe I made a mistake in the setup.Wait, let me try another approach.Let me denote the original speed as ( r ) mph.The circumference is 15 feet, which is ( frac{15}{5280} ) miles.Time for one rotation in hours is ( t = frac{text{distance}}{text{speed}} = frac{15/5280}{r} ) hours.Convert that to seconds: ( t = frac{15}{5280 r} times 3600 = frac{15 times 3600}{5280 r} = frac{54000}{5280 r} = frac{54000 ÷ 240}{5280 ÷ 240 r} = frac{225}{22 r} ) seconds.So, original time per rotation is ( t = frac{225}{22 r} ) seconds.Now, when the time is reduced by ( frac{1}{3} ) seconds, the new time is ( t - frac{1}{3} ), and the new speed is ( r + 6 ) mph.So, the new time is also equal to ( frac{225}{22 (r + 6)} ) seconds.Therefore, we have:( frac{225}{22 r} - frac{1}{3} = frac{225}{22 (r + 6)} ).Let me write this equation:( frac{225}{22 r} - frac{1}{3} = frac{225}{22 (r + 6)} ).Let me subtract ( frac{225}{22 (r + 6)} ) from both sides:( frac{225}{22 r} - frac{225}{22 (r + 6)} - frac{1}{3} = 0 ).Factor out ( frac{225}{22} ):( frac{225}{22} left( frac{1}{r} - frac{1}{r + 6} right) - frac{1}{3} = 0 ).Compute ( frac{1}{r} - frac{1}{r + 6} = frac{(r + 6) - r}{r(r + 6)} = frac{6}{r(r + 6)} ).So, substituting back:( frac{225}{22} times frac{6}{r(r + 6)} - frac{1}{3} = 0 ).Simplify ( frac{225}{22} times 6 = frac{1350}{22} = frac{675}{11} ).So, ( frac{675}{11} times frac{1}{r(r + 6)} - frac{1}{3} = 0 ).Let me write this as:( frac{675}{11 r(r + 6)} = frac{1}{3} ).Multiply both sides by ( 11 r(r + 6) ):( 675 = frac{11 r(r + 6)}{3} ).Multiply both sides by 3:( 2025 = 11 r(r + 6) ).Expand the right side:( 2025 = 11 r^2 + 66 r ).Bring all terms to one side:( 11 r^2 + 66 r - 2025 = 0 ).Divide the entire equation by 11 to simplify:( r^2 + 6 r - 184.0909 ≈ 0 ).Wait, 2025 ÷ 11 is approximately 184.0909.But maybe I should keep it as fractions.2025 ÷ 11 = 184.0909, but let me see:2025 = 11 r^2 + 66 r.So, 11 r^2 + 66 r - 2025 = 0.Let me try to factor this or use quadratic formula.Quadratic formula: ( r = frac{-66 pm sqrt{66^2 - 4 times 11 times (-2025)}}{2 times 11} ).Compute discriminant:( 66^2 = 4356 ).( 4 times 11 times 2025 = 44 times 2025 = 89100 ).So, discriminant is ( 4356 + 89100 = 93456 ).sqrt(93456). Let me see, 306^2 = 93636, which is higher. 305^2 = 93025, which is lower. So, sqrt(93456) is between 305 and 306.Compute 305.5^2 = (305 + 0.5)^2 = 305^2 + 2*305*0.5 + 0.5^2 = 93025 + 305 + 0.25 = 93330.25.Still lower than 93456.305.7^2: Let me compute 305.7^2.305^2 = 93025.2*305*0.7 = 427.0.7^2 = 0.49.So, 305.7^2 = 93025 + 427 + 0.49 = 93452.49.Close to 93456. Difference is 93456 - 93452.49 = 3.51.So, sqrt(93456) ≈ 305.7 + 3.51/(2*305.7) ≈ 305.7 + 0.0057 ≈ 305.7057.So, approximately 305.706.Thus, ( r = frac{-66 pm 305.706}{22} ).We discard the negative solution, so:( r = frac{-66 + 305.706}{22} = frac{239.706}{22} ≈ 10.8957 ) mph.So, approximately 10.896 mph, which is about 10.9 mph.But the options are 8, 9, 10, 11. So, 10.9 is closer to 11, but let me check if I did everything correctly.Wait, maybe I made a mistake in the quadratic setup.Let me go back.We had:( frac{675}{11 r(r + 6)} = frac{1}{3} ).Multiply both sides by ( 11 r(r + 6) ):( 675 = frac{11 r(r + 6)}{3} ).Multiply both sides by 3:( 2025 = 11 r(r + 6) ).So, ( 11 r^2 + 66 r - 2025 = 0 ).Yes, that's correct.Quadratic formula: ( r = frac{-66 pm sqrt{66^2 + 4 times 11 times 2025}}{2 times 11} ).Which is ( r = frac{-66 pm sqrt{4356 + 89100}}{22} = frac{-66 pm sqrt{93456}}{22} ≈ frac{-66 + 305.706}{22} ≈ 10.896 ).So, approximately 10.896 mph.But the options are 8, 9, 10, 11. So, 10.896 is closest to 11, but maybe the exact value is 9.Wait, let me check if r = 9 satisfies the equation.Let me plug r = 9 into the equation:( 11 r^2 + 66 r - 2025 = 11*81 + 66*9 - 2025 = 891 + 594 - 2025 = 1485 - 2025 = -540 ≠ 0 ).So, r = 9 is not a solution.Wait, maybe I made a mistake in the setup.Let me try another approach.Let me denote the original speed as ( r ) mph.The circumference is 15 feet, which is ( frac{15}{5280} ) miles.Time for one rotation in hours is ( t = frac{text{distance}}{text{speed}} = frac{15/5280}{r} ) hours.Convert that to seconds: ( t = frac{15}{5280 r} times 3600 = frac{15 times 3600}{5280 r} = frac{54000}{5280 r} = frac{54000 ÷ 240}{5280 ÷ 240 r} = frac{225}{22 r} ) seconds.So, original time per rotation is ( t = frac{225}{22 r} ) seconds.When the time is reduced by ( frac{1}{3} ) seconds, the new time is ( t - frac{1}{3} ), and the new speed is ( r + 6 ) mph.So, the new time is also equal to ( frac{225}{22 (r + 6)} ) seconds.Therefore, we have:( frac{225}{22 r} - frac{1}{3} = frac{225}{22 (r + 6)} ).Let me write this equation:( frac{225}{22 r} - frac{225}{22 (r + 6)} = frac{1}{3} ).Factor out ( frac{225}{22} ):( frac{225}{22} left( frac{1}{r} - frac{1}{r + 6} right) = frac{1}{3} ).Compute ( frac{1}{r} - frac{1}{r + 6} = frac{(r + 6) - r}{r(r + 6)} = frac{6}{r(r + 6)} ).So, substituting back:( frac{225}{22} times frac{6}{r(r + 6)} = frac{1}{3} ).Simplify ( frac{225}{22} times 6 = frac{1350}{22} = frac{675}{11} ).So, ( frac{675}{11} times frac{1}{r(r + 6)} = frac{1}{3} ).Multiply both sides by ( r(r + 6) ):( frac{675}{11} = frac{r(r + 6)}{3} ).Multiply both sides by 3:( frac{2025}{11} = r(r + 6) ).So, ( r^2 + 6r - frac{2025}{11} = 0 ).Multiply both sides by 11 to eliminate the fraction:( 11 r^2 + 66 r - 2025 = 0 ).Which is the same quadratic equation as before.So, solving this quadratic equation, we get ( r ≈ 10.896 ) mph.But since the options are 8, 9, 10, 11, and 10.896 is closest to 11, but let me check if 9 is a solution.Wait, when I plugged r = 9, it didn't satisfy the equation. So, maybe the answer is 9, but my calculations say 10.896.Wait, perhaps I made a mistake in the unit conversion.Let me double-check the unit conversion.Original speed: ( r ) mph.Circumference: 15 feet.Time per rotation: ( t = frac{15}{r times 5280/3600} ) seconds.Wait, let me re-express this.Speed in feet per second: ( r times frac{5280}{3600} = r times frac{22}{15} ) ft/s.So, time per rotation: ( t = frac{15}{r times frac{22}{15}} = frac{15 times 15}{22 r} = frac{225}{22 r} ) seconds. That's correct.So, when the time is reduced by ( frac{1}{3} ) seconds, the new time is ( t - frac{1}{3} ), and the new speed is ( r + 6 ) mph.So, new time: ( t - frac{1}{3} = frac{225}{22 (r + 6)} ).Thus, equation:( frac{225}{22 r} - frac{1}{3} = frac{225}{22 (r + 6)} ).Which leads to the quadratic equation as before.So, the solution is approximately 10.896 mph, which is closest to 11, but the options include 10 and 11.Wait, maybe I should check if r = 9 satisfies the original equation.Let me plug r = 9 into the equation:( frac{225}{22 * 9} - frac{1}{3} = frac{225}{22 * (9 + 6)} ).Compute left side:( frac{225}{198} - frac{1}{3} ≈ 1.136 - 0.333 ≈ 0.803 ).Right side:( frac{225}{22 * 15} = frac{225}{330} ≈ 0.6818 ).So, 0.803 ≈ 0.6818? No, not equal. So, r = 9 is not a solution.Similarly, try r = 10:Left side:( frac{225}{220} - frac{1}{3} ≈ 1.0227 - 0.333 ≈ 0.6897 ).Right side:( frac{225}{22 * 16} = frac{225}{352} ≈ 0.639 ).Not equal.r = 11:Left side:( frac{225}{242} - frac{1}{3} ≈ 0.9298 - 0.333 ≈ 0.5968 ).Right side:( frac{225}{22 * 17} = frac{225}{374} ≈ 0.5997 ).These are very close: 0.5968 vs 0.5997. So, r = 11 is very close to satisfying the equation.Wait, but my quadratic solution gave r ≈ 10.896, which is very close to 11.But the options are 8, 9, 10, 11.Given that 10.896 is very close to 11, and when I plug r = 11, the left and right sides are almost equal, I think the intended answer is 9, but my calculations say 11.Wait, maybe I made a mistake in the setup.Wait, let me try to solve the quadratic equation exactly.We have ( 11 r^2 + 66 r - 2025 = 0 ).Let me try to factor this.Looking for two numbers that multiply to ( 11 * (-2025) = -22275 ) and add to 66.Hmm, 22275 is a large number. Let me see, 22275 ÷ 75 = 297. So, 75 and 297. 75 + 297 = 372, which is not 66.Alternatively, 22275 ÷ 25 = 891. 25 and 891 add to 916, not helpful.Alternatively, 22275 ÷ 15 = 1485. 15 and 1485 add to 1500, not helpful.Alternatively, 22275 ÷ 5 = 4455. 5 and 4455 add to 4460, not helpful.Alternatively, 22275 ÷ 3 = 7425. 3 and 7425 add to 7428, not helpful.Hmm, seems like it's not factorable easily. So, I think the quadratic formula is the way to go.So, ( r = frac{-66 pm sqrt{66^2 + 4 times 11 times 2025}}{2 times 11} = frac{-66 pm sqrt{4356 + 89100}}{22} = frac{-66 pm sqrt{93456}}{22} ).As before, sqrt(93456) ≈ 305.706.So, ( r ≈ frac{-66 + 305.706}{22} ≈ frac{239.706}{22} ≈ 10.896 ).So, approximately 10.896 mph.But since the options are 8, 9, 10, 11, and 10.896 is very close to 11, I think the intended answer is 9, but my calculations say 11.Wait, maybe I made a mistake in the setup.Wait, let me try to solve the problem using another approach.Let me denote the original speed as ( r ) mph.The circumference is 15 feet, which is ( frac{15}{5280} ) miles.Time for one rotation in hours is ( t = frac{text{distance}}{text{speed}} = frac{15/5280}{r} ) hours.Convert that to seconds: ( t = frac{15}{5280 r} times 3600 = frac{15 times 3600}{5280 r} = frac{54000}{5280 r} = frac{54000 ÷ 240}{5280 ÷ 240 r} = frac{225}{22 r} ) seconds.So, original time per rotation is ( t = frac{225}{22 r} ) seconds.When the time is reduced by ( frac{1}{3} ) seconds, the new time is ( t - frac{1}{3} ), and the new speed is ( r + 6 ) mph.So, the new time is also equal to ( frac{225}{22 (r + 6)} ) seconds.Therefore, we have:( frac{225}{22 r} - frac{1}{3} = frac{225}{22 (r + 6)} ).Let me write this equation:( frac{225}{22 r} - frac{225}{22 (r + 6)} = frac{1}{3} ).Factor out ( frac{225}{22} ):( frac{225}{22} left( frac{1}{r} - frac{1}{r + 6} right) = frac{1}{3} ).Compute ( frac{1}{r} - frac{1}{r + 6} = frac{6}{r(r + 6)} ).So, substituting back:( frac{225}{22} times frac{6}{r(r + 6)} = frac{1}{3} ).Simplify ( frac{225}{22} times 6 = frac{1350}{22} = frac{675}{11} ).So, ( frac{675}{11} times frac{1}{r(r + 6)} = frac{1}{3} ).Multiply both sides by ( r(r + 6) ):( frac{675}{11} = frac{r(r + 6)}{3} ).Multiply both sides by 3:( frac{2025}{11} = r(r + 6) ).So, ( r^2 + 6r - frac{2025}{11} = 0 ).Multiply both sides by 11:( 11 r^2 + 66 r - 2025 = 0 ).Quadratic formula: ( r = frac{-66 pm sqrt{66^2 + 4 times 11 times 2025}}{2 times 11} = frac{-66 pm sqrt{4356 + 89100}}{22} = frac{-66 pm sqrt{93456}}{22} ).As before, sqrt(93456) ≈ 305.706.So, ( r ≈ frac{-66 + 305.706}{22} ≈ 10.896 ) mph.So, the original speed is approximately 10.896 mph, which is closest to 11 mph.But the options are 8, 9, 10, 11. So, the answer should be 11.Wait, but earlier when I plugged r = 11, the left and right sides were almost equal, which supports this.So, I think the correct answer is 11 mph, which is option D.But the initial quadratic solution gave me approximately 10.896, which is very close to 11.Therefore, the original speed was 11 mph.

💡Okay, so I have this problem where I need to evaluate some integrals involving ( x^n e^{-ax} ) from 0 to t, and then find their limits as t approaches infinity. Let me try to break this down step by step.First, for part (1), I need to evaluate ( I_1(t) ). That means I have to compute the integral ( int_0^t x e^{-ax} dx ). Hmm, I remember that integration by parts is useful for integrals involving products of functions, especially when one part can be easily integrated or differentiated. So, let me recall the formula for integration by parts: ( int u dv = uv - int v du ).Let me set ( u = x ) and ( dv = e^{-ax} dx ). Then, I need to find ( du ) and ( v ). Differentiating ( u = x ) gives ( du = dx ). Integrating ( dv = e^{-ax} dx ) gives ( v = -frac{1}{a} e^{-ax} ). Okay, so plugging these into the integration by parts formula:( I_1(t) = uv - int v du = left[ -frac{x}{a} e^{-ax} right]_0^t - int_0^t left( -frac{1}{a} e^{-ax} right) dx ).Simplifying the boundary terms first: at ( x = t ), it's ( -frac{t}{a} e^{-at} ), and at ( x = 0 ), it's ( -frac{0}{a} e^{0} = 0 ). So, the first term becomes ( -frac{t}{a} e^{-at} ).Now, the remaining integral is ( - int_0^t -frac{1}{a} e^{-ax} dx = frac{1}{a} int_0^t e^{-ax} dx ). Integrating ( e^{-ax} ) gives ( -frac{1}{a} e^{-ax} ), so evaluating from 0 to t:( frac{1}{a} left[ -frac{1}{a} e^{-ax} right]_0^t = frac{1}{a} left( -frac{1}{a} e^{-at} + frac{1}{a} right) = frac{1}{a^2} (1 - e^{-at}) ).Putting it all together, ( I_1(t) = -frac{t}{a} e^{-at} + frac{1}{a^2} (1 - e^{-at}) ). That seems right. Maybe I can factor out ( e^{-at} ) for simplicity:( I_1(t) = frac{1}{a^2} - left( frac{t}{a} + frac{1}{a^2} right) e^{-at} ).Okay, that looks good. So that's part (1) done.Moving on to part (2), I need to find a relation between ( I_{n+1}(t) ) and ( I_n(t) ). This probably involves using integration by parts again, similar to what I did in part (1). Let me try that.So, ( I_{n+1}(t) = int_0^t x^{n+1} e^{-ax} dx ). Let me set ( u = x^{n+1} ) and ( dv = e^{-ax} dx ). Then, ( du = (n+1) x^n dx ) and ( v = -frac{1}{a} e^{-ax} ).Applying integration by parts:( I_{n+1}(t) = uv - int v du = left[ -frac{x^{n+1}}{a} e^{-ax} right]_0^t - int_0^t left( -frac{1}{a} e^{-ax} right) (n+1) x^n dx ).Simplifying the boundary terms: at ( x = t ), it's ( -frac{t^{n+1}}{a} e^{-at} ), and at ( x = 0 ), it's 0 since ( x^{n+1} ) is zero. So, the first term is ( -frac{t^{n+1}}{a} e^{-at} ).The remaining integral is ( (n+1)/a int_0^t x^n e^{-ax} dx = frac{n+1}{a} I_n(t) ).Putting it all together:( I_{n+1}(t) = -frac{t^{n+1}}{a} e^{-at} + frac{n+1}{a} I_n(t) ).So, that's the relation between ( I_{n+1}(t) ) and ( I_n(t) ). Good.Now, part (3) asks me to prove by induction that the limit ( lim_{t to infty} I_n(t) ) exists for all natural numbers ( n ). They also note that I can use ( lim_{t to infty} t^n e^{-at} = 0 ) without proof.Alright, so induction is a good approach here. Let me recall how induction works: I need to prove the base case, then assume it's true for some ( n ), and then show it's true for ( n+1 ).First, the base case: ( n = 1 ). From part (1), I have ( I_1(t) = frac{1}{a^2} - left( frac{t}{a} + frac{1}{a^2} right) e^{-at} ). Taking the limit as ( t to infty ), the term ( left( frac{t}{a} + frac{1}{a^2} right) e^{-at} ) goes to zero because ( e^{-at} ) decays exponentially and dominates the polynomial terms. So, ( lim_{t to infty} I_1(t) = frac{1}{a^2} ). So, the base case holds.Now, assume that for some ( n = k ), ( lim_{t to infty} I_k(t) ) exists. Let's denote this limit as ( L_k ). Then, we need to show that ( lim_{t to infty} I_{k+1}(t) ) exists.From part (2), we have:( I_{k+1}(t) = -frac{t^{k+1}}{a} e^{-at} + frac{k+1}{a} I_k(t) ).Taking the limit as ( t to infty ):( lim_{t to infty} I_{k+1}(t) = lim_{t to infty} left( -frac{t^{k+1}}{a} e^{-at} + frac{k+1}{a} I_k(t) right) ).We know that ( lim_{t to infty} t^{k+1} e^{-at} = 0 ), so the first term is zero. The second term is ( frac{k+1}{a} lim_{t to infty} I_k(t) = frac{k+1}{a} L_k ). Therefore, ( lim_{t to infty} I_{k+1}(t) = frac{k+1}{a} L_k ), which exists because ( L_k ) exists by the induction hypothesis.Thus, by induction, ( lim_{t to infty} I_n(t) ) exists for all natural numbers ( n ). That completes part (3).Finally, part (4) asks me to find ( lim_{t to infty} I_n(t) ). From part (3), we know the limit exists, and from the relation in part (2), we can express ( L_{n+1} = frac{n+1}{a} L_n ), where ( L_n = lim_{t to infty} I_n(t) ).This recurrence relation suggests that each ( L_{n+1} ) is related to ( L_n ) by a factor of ( frac{n+1}{a} ). Let me try to find a general expression for ( L_n ).Starting from the base case:( L_1 = frac{1}{a^2} ).Using the recurrence:( L_2 = frac{2}{a} L_1 = frac{2}{a} cdot frac{1}{a^2} = frac{2}{a^3} ).Similarly,( L_3 = frac{3}{a} L_2 = frac{3}{a} cdot frac{2}{a^3} = frac{6}{a^4} ).Wait, that's ( 3! / a^{4} ). Hmm, let me see:( L_1 = 1! / a^{2} ).( L_2 = 2! / a^{3} ).( L_3 = 3! / a^{4} ).So, it seems that ( L_n = frac{n!}{a^{n+1}} ).Let me test this hypothesis with the recurrence relation. Suppose ( L_n = frac{n!}{a^{n+1}} ), then:( L_{n+1} = frac{n+1}{a} L_n = frac{n+1}{a} cdot frac{n!}{a^{n+1}} = frac{(n+1)!}{a^{n+2}} ).Which matches the pattern. Therefore, by induction, ( L_n = frac{n!}{a^{n+1}} ) for all natural numbers ( n ).So, putting it all together, the limit is ( frac{n!}{a^{n+1}} ).**Final Answer**The limit is boxed{dfrac{n!}{a^{n+1}}}.

💡Okay, so I have this problem where I need to find a point O inside triangle ABC such that the sum of the lengths OA, OB, and OC is minimized. The problem also mentions paying attention to the case when one of the angles of the triangle is greater than 120 degrees. Hmm, interesting. I remember something about Fermat points and Torricelli points in geometry, which deal with minimizing distances from a point to the vertices of a triangle. Maybe that's related?Let me start by recalling what I know. The Fermat-Torricelli point, or simply the Fermat point, is a point such that the total distance from the three vertices of the triangle to this point is minimized. If all the angles of the triangle are less than 120 degrees, the Fermat point is inside the triangle, and each of the lines from the vertices to this point makes 120-degree angles with each other. But if one of the angles is 120 degrees or more, the Fermat point coincides with the vertex of that large angle. Is that right?So, in this problem, we need to consider two cases: when all angles are less than 120 degrees and when one angle is 120 degrees or more. Let me think about each case separately.First, if all angles are less than 120 degrees, then the Fermat point is inside the triangle. To construct it, I believe you can draw equilateral triangles on the sides of the original triangle and connect their centroids or something like that. I might be mixing it up with the Napoleon points, though. Maybe I should look into the construction more carefully.Alternatively, I remember that the Fermat point can be found by constructing lines from each vertex such that each pair of lines forms a 120-degree angle. If I can construct such lines, their intersection should give me the Fermat point. But how exactly do I construct those lines?Let me try to visualize it. Suppose I have triangle ABC with all angles less than 120 degrees. I need to find a point O inside the triangle such that OA + OB + OC is minimized. If I construct an equilateral triangle on one side, say BC, and then connect the new vertex to A, the intersection of this line with the original triangle might lead me to the Fermat point. Is that correct?Wait, maybe it's better to use the method of constructing lines from each vertex at 120 degrees to each other. So, from point A, I draw a line making 120 degrees with AB, and from point B, I draw a line making 120 degrees with BC, and their intersection should be the Fermat point. That seems plausible.But I'm not entirely sure. Maybe I should try to verify this. If I consider the properties of the Fermat point, it should satisfy the condition that the angles between the lines from the point to each vertex are all 120 degrees. So, if I can ensure that each pair of lines from O to the vertices forms a 120-degree angle, then O is indeed the Fermat point.Now, what if one of the angles of the triangle is greater than or equal to 120 degrees? In that case, the Fermat point doesn't lie inside the triangle anymore. Instead, it coincides with the vertex of the obtuse angle. So, for example, if angle C is greater than or equal to 120 degrees, then the point O that minimizes OA + OB + OC is just point C itself.Why is that? Well, if angle C is large, say 120 degrees or more, then trying to place a point inside the triangle to minimize the sum of distances might not be beneficial because the large angle already restricts the possible positions. By placing O at C, we can potentially reduce the total distance since moving O away from C would increase the distances to the other vertices more than it would decrease the distance to C.Let me try to think of a specific example. Suppose I have a triangle where angle C is 150 degrees. If I try to place O somewhere inside the triangle, the distances OA and OB would be quite large because of the wide angle at C. On the other hand, if I place O at C, then OC is zero, and OA and OB would just be the lengths of sides AC and BC, which might actually be shorter than trying to have O somewhere else.Hmm, that makes sense. So, in cases where one angle is 120 degrees or more, the optimal point O is just the vertex of that large angle. But when all angles are less than 120 degrees, we need to construct the Fermat point inside the triangle.I wonder if there's a way to formally prove why the Fermat point minimizes the sum of distances. Maybe using calculus or some geometric principles. I recall that in optimization problems, especially in geometry, reflection properties often come into play.For instance, in the case of the Fermat point, reflecting the triangle across one of its sides and then finding a straight line path might help in determining the minimal sum of distances. This is similar to how light reflects off a surface, taking the path of least time, which is analogous to minimizing distance.Let me try to elaborate. If I reflect point A over the line BC, I get a new point A'. Then, the minimal path from A to B to C would correspond to a straight line from A' to C, intersecting BC at some point. But I'm not sure if this directly applies to the Fermat point. Maybe I need to reflect multiple times or use a different approach.Alternatively, I remember that the Fermat point can be found as the intersection of the lines constructed by rotating the sides of the triangle by 60 degrees. So, if I take side AB and rotate it 60 degrees around point A, and similarly rotate side AC 60 degrees around point C, their intersection should give me the Fermat point. That sounds complicated, but perhaps it's a systematic way to construct it.Let me try to sketch this out mentally. If I have triangle ABC, and I rotate side AB 60 degrees around point A, I get a new line. Similarly, rotating side AC 60 degrees around point C gives another line. The intersection of these two lines should be the Fermat point O. Then, connecting O to A, B, and C should give me the minimal sum of distances.But how do I know this is the case? Maybe by considering the properties of equilateral triangles and the angles involved. If each rotation is 60 degrees, then the angles between OA, OB, and OC should naturally be 120 degrees, which is consistent with the definition of the Fermat point.I think I'm getting closer. So, in summary, for a triangle with all angles less than 120 degrees, the Fermat point is inside the triangle, and it's constructed by rotating the sides by 60 degrees and finding their intersection. For a triangle with an angle of 120 degrees or more, the Fermat point coincides with the vertex of that large angle.But I still feel like I need a more rigorous proof or explanation. Maybe I can use calculus to minimize the sum of distances. Let's denote the coordinates of points A, B, and C as (x_A, y_A), (x_B, y_B), and (x_C, y_C), respectively. Then, the sum of distances OA + OB + OC can be written as:OA + OB + OC = sqrt((x - x_A)^2 + (y - y_A)^2) + sqrt((x - x_B)^2 + (y - y_B)^2) + sqrt((x - x_C)^2 + (y - y_C)^2)To find the minimum, I can take the partial derivatives with respect to x and y, set them equal to zero, and solve for x and y. However, this seems quite involved because the square roots make the derivatives messy.Alternatively, maybe I can use geometric transformations or properties. For example, using vectors or complex numbers to represent the points and then finding the point O that minimizes the sum of the magnitudes of the vectors from O to each vertex.Wait, another idea: using the concept of geometric medians. The geometric median minimizes the sum of distances to a set of given points. In the case of three points forming a triangle, the geometric median is the Fermat-Torricelli point. So, perhaps I can use properties of the geometric median to find O.But I'm not sure about the exact properties or how to apply them here. Maybe I should look into the Weiszfeld algorithm, which is an iterative method to find the geometric median. However, that might be more computational than what is needed here.Going back to the geometric construction, I think the key lies in the 120-degree angles. If I can ensure that the angles between OA, OB, and OC are all 120 degrees, then O is the Fermat point. So, how can I construct such a point?One method I remember is constructing equilateral triangles on the sides of the original triangle and then connecting their centroids. The intersection of these lines should give the Fermat point. Let me try to visualize this.Suppose I construct an equilateral triangle on side AB, outside of triangle ABC, and let its centroid be G1. Similarly, construct an equilateral triangle on side BC, outside of ABC, with centroid G2, and another on side AC, outside of ABC, with centroid G3. Then, the lines connecting G1 to C, G2 to A, and G3 to B should intersect at the Fermat point O.But I'm not entirely sure if this is accurate. Maybe it's better to use the method of rotating the triangle. If I rotate triangle ABC by 60 degrees around one of its vertices, say A, then the image of point B will be somewhere, and the line connecting this image to C should intersect the original triangle at the Fermat point.This seems a bit abstract, but perhaps it's a way to construct the point. Alternatively, I can use the property that the Fermat point is the common intersection point of the three lines, each making 120-degree angles with the sides of the triangle.Wait, I think I need to clarify this. The Fermat point is such that each of the lines from the point to the vertices makes 120-degree angles with each other, not necessarily with the sides of the triangle. So, the angles between OA, OB, and OC are each 120 degrees.This is a crucial distinction. So, the angles between the segments OA, OB, and OC are 120 degrees, not the angles between OA and the sides of the triangle. That makes more sense.So, to construct the Fermat point, I need to ensure that the angles between OA, OB, and OC are all 120 degrees. How can I achieve this?One way is to use the method of constructing equilateral triangles on the sides and then connecting their vertices. For example, construct an equilateral triangle on side BC, then connect its new vertex to A. Similarly, construct an equilateral triangle on side AC and connect its new vertex to B. The intersection of these two lines should be the Fermat point.Let me try to think through this step-by-step. Suppose I have triangle ABC. I construct an equilateral triangle BCD on side BC, outside of ABC. Then, I connect point D to point A. Similarly, I construct an equilateral triangle ACE on side AC, outside of ABC, and connect point E to point B. The intersection of lines AD and BE should be the Fermat point O.Is this correct? I think so, because by constructing equilateral triangles, we're effectively creating 60-degree angles, which when connected back to the opposite vertices, should result in the 120-degree angles at point O.But I need to verify this. Let me consider the angles involved. If I have an equilateral triangle BCD, then angle BCD is 60 degrees. When I connect D to A, the angle between AD and BC should be such that when combined with the angle at O, it results in 120 degrees.Hmm, maybe I'm overcomplicating it. Perhaps it's better to use vector geometry or coordinate geometry to find the exact coordinates of O.Let me assign coordinates to the triangle. Suppose I place point A at (0, 0), point B at (b, 0), and point C at (c, d). Then, I can express the distances OA, OB, and OC in terms of the coordinates of O, say (x, y). The sum OA + OB + OC would then be:sqrt(x^2 + y^2) + sqrt((x - b)^2 + y^2) + sqrt((x - c)^2 + (y - d)^2)To minimize this sum, I can take partial derivatives with respect to x and y, set them to zero, and solve for x and y. However, as I thought earlier, this leads to a system of nonlinear equations which might not have a closed-form solution.Alternatively, maybe I can use the method of Lagrange multipliers, but that also seems complicated. Perhaps there's a geometric approach that avoids calculus.Wait, another idea: using the concept of reflection. If I reflect point O across the sides of the triangle, the minimal path from A to B to C via O should correspond to a straight line in the reflected plane. But I'm not sure how to apply this here.Let me think again about the Fermat point. If all angles are less than 120 degrees, the Fermat point is inside the triangle, and each of the lines from the vertices to this point makes 120-degree angles with each other. So, if I can construct such a point, that should be the minimizer.But how do I construct it? Maybe by using the method of rotating the triangle. For example, if I rotate triangle ABC by 60 degrees around point A, the image of point B will be somewhere, and the line connecting this image to C should intersect the original triangle at the Fermat point.Let me try to elaborate. Suppose I rotate point B around point A by 60 degrees to get point B'. Then, the line connecting B' to C should intersect the original triangle at the Fermat point O. Similarly, if I rotate point C around point B by 60 degrees to get point C', the line connecting C' to A should also pass through O. The intersection of these two lines should give me the Fermat point.This seems like a plausible method. By rotating the triangle, I'm effectively creating a new point that, when connected, forms the necessary 120-degree angles at O. So, the intersection of these two lines (B'C and C'A) should be the Fermat point.But I need to make sure that this construction actually results in the Fermat point. Let me consider the angles involved. If I rotate point B around A by 60 degrees, the angle between AB and AB' is 60 degrees. Then, the line B'C should form a 120-degree angle with AO, because the rotation effectively adds a 60-degree angle, and the original angle at A is less than 120 degrees.Hmm, I think I'm getting somewhere. So, by constructing these rotated points and connecting them, I can find the Fermat point where the angles between OA, OB, and OC are all 120 degrees.But what if one of the angles is 120 degrees or more? In that case, as I thought earlier, the Fermat point coincides with the vertex of the large angle. So, if angle C is 120 degrees or more, then point O should be at point C.Why is that? Well, if angle C is large, say 150 degrees, then trying to place O inside the triangle would result in OA and OB being quite long because of the wide angle. On the other hand, placing O at C would make OC zero, and OA and OB would just be the lengths of sides AC and BC, which might actually be shorter than trying to have O somewhere else inside the triangle.Let me test this with a specific example. Suppose I have triangle ABC with angle C = 150 degrees, and sides AC = 1, BC = 1, and AB = sqrt(3). If I place O at C, then OA + OB + OC = 1 + 1 + 0 = 2. If I try to place O somewhere inside the triangle, say at the centroid, then OA, OB, and OC would each be something like 0.5, but the sum would still be around 1.5, which is less than 2. Wait, that contradicts my earlier thought.Hmm, maybe I made a mistake. If angle C is 150 degrees, and sides AC and BC are both 1, then the triangle is quite "stretched" at C. The centroid would be located closer to C, but the distances OA, OB, and OC might actually be longer than just placing O at C.Wait, no, the centroid divides the medians in a 2:1 ratio, so the distances from the centroid to the vertices are actually longer than the distances from the vertex itself. So, in this case, placing O at C would indeed give a smaller sum of distances.Let me calculate it properly. If O is at C, then OA = AC = 1, OB = BC = 1, and OC = 0, so the total is 2. If O is at the centroid, then the distances OA, OB, and OC would each be something like sqrt((2/3)^2 + (1/3)^2) ≈ 0.745, so the total would be around 2.236, which is larger than 2. So, placing O at C gives a smaller sum.Therefore, when one angle is 120 degrees or more, placing O at that vertex minimizes the sum of distances. That makes sense because the large angle makes it unfavorable to have O inside the triangle, as the distances to the other vertices would be too large.So, to summarize, if all angles of triangle ABC are less than 120 degrees, the point O that minimizes OA + OB + OC is the Fermat-Torricelli point inside the triangle, where each of the lines from O to the vertices makes 120-degree angles with each other. If one of the angles is 120 degrees or more, then O coincides with the vertex of that large angle.But I still feel like I need a more rigorous proof or explanation. Maybe I can use the concept of geometric medians or some optimization principles. However, given the time constraints, I think I've covered the main ideas and can now present the step-by-step explanation.

💡To find the derivative of the function ( y = 2x ), I start by recalling the basic rules of differentiation.The function is a linear function where the variable ( x ) is multiplied by a constant coefficient of 2. According to the power rule of differentiation, the derivative of ( x ) with respect to ( x ) is 1. Multiplying this result by the constant coefficient 2 gives the derivative of the entire function.Therefore, the derivative ( y' ) is 2.