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Find the length of the curve with parametric equations π₯ equals π to the power of π‘ minus π‘ and π¦ equals four π to the power of π‘ over two, where π‘ is greater than or equal to zero and less than or equal to two.
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We know that the formula we used to find the arc length of curves defined parametrically from values of π‘ greater than or equal to πΌ and less than or equal to π½ is the definite integral between πΌ and π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to π‘.
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Now, in this case, weβre interested in the length of the curve, where π‘ is greater than or equal to zero and less than or equal to π‘.
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So weβll let πΌ to be equal to zero and π½ be equal to two.
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Then, our parametric equations are π₯ equals π to the power of π‘ minus π‘ and π¦ equals four π to the power of π‘ over two.
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Itβs quite clear that weβre going to need to work out dπ₯ by dπ‘ and dπ¦ by dπ‘.
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And so, we firstly recall that the derivative of π to the power of π‘ is π to the power of π‘.
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The derivative of negative π‘ is one.
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So dπ₯ by dπ‘ is π to the power of π‘ minus one.
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Now, weβre going to use the chain rule to differentiate π¦ with respect to π‘.
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We let π’ be equal to π‘ over two.
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So that dπ’ by dπ‘ is equal to one-half.
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Then, dπ¦ by dπ‘ is dπ¦ by dπ’ times dπ’ by dπ‘.
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Now, π¦ is equal to four π to the power of π’.
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So dπ¦ by dπ‘ is four π to the power of π’ times a half, which is two π to the power of π’.
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But of course, we want dπ¦ by dπ‘ in terms of π‘.
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So we replace π’ with π‘ over two.
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And we find that dπ¦ by dπ‘ equals two π to the power of π‘ over two.
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Now, in fact, for our arc length formula, we need dπ₯ by dπ‘ squared and dπ¦ by dπ‘ squared.
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So weβre going to square each of our expressions.
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When we do, we find that π to the power of π‘ minus one squared is π to the power of two π‘ minus two π to the power of π‘ plus one.
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And dπ¦ by dπ‘ squared is simply four π to the power of π‘.
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Now, we substitute everything we know into our formula for the arc length.
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And we get the definite integral between zero and two of the square root of π to the power of two π‘ minus two π to the power of π‘ plus one plus four π to the power of π‘ with respect to π‘.
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We notice that negative two π to the power of π‘ plus four π to the power of π‘ is two π to the power of π‘.
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And thatβs great because we see we can factor π to the power of two π‘ plus two π to the power of π‘ plus one a little like we would with a quadratic.
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We get π to the power of π‘ plus one times π to the power of π‘ plus one or π to the power of π‘ plus one squared.
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And of course, the square root of π to the power of π‘ plus one squared is just π to the power of π‘ plus one.
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When we integrate π to the power of π‘, we get π to the power of π‘.
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And the integral of one is π‘.
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This means the arc length is π squared plus two minus π to the power of of zero plus zero.
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And, of course, π to the power of zero is one.
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So this simplifies to π to the power of two plus one.
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And the length of the curve with our parametric equations for values of π‘ from zero to two is π to the power of two plus one units.