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Find the length of the curve with parametric equations π₯ is equal to π‘ squared and π¦ is equal to one-third π‘ cubed, where π‘ is greater than or equal to zero and π‘ is less than or equal to one.
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The question wants us to find the length of a curve defined by a pair of parametric equations.
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And we recall we can calculate the length of a curve defined as a pair of parametric equations π₯ is equal to π of π‘ and π¦ is equal to π of π‘, where π‘ is greater than or equal to πΌ and π‘ is less than or equal to π½.
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As the integral from πΌ to π½ of the square root of π prime of π‘ squared plus π prime of π‘ squared with respect to π‘.
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In our question, we have π₯ is equal to π‘ squared and π¦ is equal to one-third π‘ cubed.
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So we set π of π‘ equal to π‘ squared and π of π‘ equal to one-third π‘ cubed.
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And we want the length of the curve where π‘ is between zero and one.
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So weβll set πΌ equal to zero and π½ equal to one.
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So the length of our parametric equation is given by the integral from zero to one of the square root of π prime of π‘ squared plus π prime of π‘ squared with respect to π‘.
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Now, we need to calculate our functions π prime of π‘ and π prime of π‘.
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We can do this by using the power rule for differentiation.
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We multiply by our exponent and then reduce the exponent by one.
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This gives us that π prime of π‘ is equal to two π‘ and π prime of π‘ is equal to π‘ squared.
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This gives us that the length of our curve is equal to the integral from zero to one of the square root of two π‘ squared plus π‘ squared squared with respect to π‘.
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We can simplify this by noticing that two π‘ all squared is equal to four π‘ squared and π‘ squared all squared is equal to π‘ to the fourth power.
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This gives us the integral from zero to one of the square root of four π‘ squared plus π‘ to the fourth power with respect to π‘.
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We see that both terms inside of our square root share a factor of π‘ squared.
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So we can rewrite this as π‘ squared multiplied by four plus π‘ squared.
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This gives us the integral from zero to one of the square root of π‘ squared multiplied by four plus π‘ squared with respect to π‘.
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And we can evaluate the square root of π‘ squared.
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Itβs just equal to π‘.
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So we need to evaluate the integral from zero to one of π‘ times the square root of four plus π‘ squared with respect to π‘.
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To integrate this, we need to notice that our innermost function, four plus π‘ squared, has the derivative of two π‘.
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But we see weβre multiplying by π‘, which is a linear multiple of two π‘.
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This tells us we can evaluate this integral by using the substitution π’ is equal to four plus π‘ squared.
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If we have that π’ is equal to four plus π‘ squared, differentiating both sides of this equation with respect to π‘ gives us that dπ’ by dπ‘ is equal to two π‘.
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And although dπ’ by dπ‘ is definitely not a fraction, when weβre using integration by substitution, it does behave a little bit like a fraction.
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We can then use this to get the equivalent statement that dπ’ is equal to two π‘ dπ‘.
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And since weβre using substitution on a definite integral, we need to rewrite the bounds of our integral.
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Weβll start with the lower bound.
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When π‘ is equal to zero, we have that π’ is equal to four plus zero squared, which is just equal to four.
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And in our upper bound, when π‘ is equal to one, we have that π’ is equal to four plus one squared, which is of course just equal to five.
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So by using the substitution π’ is equal to four plus π‘ squared, we have our integral is equal to the integral from four to five.
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We know that dπ’ is equal to two π‘ dπ‘.
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So π‘ dπ‘ is equal to a half dπ’.
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And using the substitution π’ is equal to four plus π‘ squared, we have the square root of four plus π‘ squared is just the square root of π’.
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So the length of our curve is equal to the integral from four to five of a half times the square root of π’ with respect to π’.
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Now, we can evaluate this integral by noticing the square root of π’ is equal to π’ to the power of a half.
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Weβll use the power rule for integration.
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Weβre now ready to evaluate this integral.
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First, we take the constant factor of a half outside of our integral.
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Then, by noticing the square root of π’ is equal to π’ to the power of a half, weβll use the power rule for integration.
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We add one to the exponent and then divide by this new exponent.
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This gives us that the length of our curve is equal to a half multiplied by two-thirds π’ to the power of three over two evaluated at the limits of our integral π’ is equal to four and π’ is equal to five.
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Evaluating this at the limits of our integral gives us a half times two-thirds of five to the power of three over two minus two-thirds of four to the power of three over two.
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We could leave our answer here, but we can simplify this further.
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First, weβll cancel the shared factor of two in our numerator and our denominator.
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Next, weβll take out the shared factor of one-third.
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This gives us one-third multiplied by five to the power of three over two minus four to the power of three over two.
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We can then simplify this further by noticing that raising a number to the power of three over two is the same as taking the square root of that number and then cubing it.
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So five to the power of three over two is equal to five root five.
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And four to the power of three over two is equal to eight.
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Therefore, weβve shown that the length of the curve of parametric equations π₯ is equal to π‘ squared and π¦ is equal to one-third π‘ cubed, where π‘ is greater than or equal to zero and π‘ is less than or equal to one.
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Is equal to five root five minus eight all divided by three.