# Video: Finding the Length of an Arc Described by a Pair of Parametric Equations between Specified Points

Find the length of the arc defined by 𝑥 = (1/2 𝑡²) and 𝑦 = (1/6) (4𝑡 + 4)^(3/2)^(3/2), from 𝑡 = 0 to 𝑡 = 2.

05:26

### Video Transcript

Find the length of the arc defined by 𝑥 is equal to one-half 𝑡 squared and 𝑦 is equal to one-sixth times four 𝑡 plus four to the power of three over two, from 𝑡 is equal to zero to 𝑡 is equal to two.

The question gives us a pair of parametric equations. And the question wants us to find the length of the arc defined by these parametric equations as 𝑡 goes from zero to two. And we recall if we’re given a pair of parametric equations 𝑥 is equal to some function 𝑓 of 𝑡 and 𝑦 is equal to some function 𝑔 of 𝑡, then we can find the length of the curve defined by this pair of parametric equations from 𝑡 is equal to 𝛼 to 𝑡 is equal to 𝛽 by evaluating the integral from 𝛼 to 𝛽 of the square root of d𝑥 by d𝑡 squared plus d𝑦 by d𝑡 squared with respect to 𝑡.

From the question, we see that we want our function 𝑓 of 𝑡 to be a half 𝑡 squared and our function 𝑔 of 𝑡 to be one-sixth times four 𝑡 plus four to the power of three over two. And since we want to find the length of the arc from 𝑡 is equal to zero to 𝑡 is equal to two, we’ll set 𝛼 equal to zero and 𝛽 equal to two. We see from our formula of the arc length we need to calculate d𝑥 by d𝑡 and d𝑦 by d𝑡. We can start by calculating d𝑥 by d𝑡. That’s the derivative of one-half 𝑡 squared with respect to 𝑡. We can differentiate this by using the power rule for differentiation. We multiply by our exponent and reduce the exponent by one. This gives us two 𝑡 to the power of two minus one divided by two. We can cancel the shared factor of two. And then we see two minus one is equal to one. So our derivative is just 𝑡.

We now want to calculate d𝑦 by d𝑡. That’s the derivative of one-sixth times four 𝑡 plus four to the power of three over two with respect to 𝑡. We can see that this is the derivative of a composition of two functions. So to differentiate this, we’re going to want to use the chain rule. We’ll set our inner function four 𝑡 plus four equal to 𝑢. The chain rule tells us if 𝑦 is some function of 𝑢 and 𝑢 is some function of 𝑡, then we can find d𝑦 by d𝑡 by finding d𝑦 by d𝑢 and multiplying this by d𝑢 by d𝑡. So by using the chain rule, we have that our derivative is equal to the derivative of one-sixth 𝑢 to the power of three over two with respect to 𝑢 multiplied by d𝑢 by d𝑡. We can evaluate the derivative of one-sixth times 𝑢 to the power of three over two with respect to 𝑢 by using the power rule for differentiation.

We multiply by our exponent and reduce the exponent by one. This gives us three over two times one-sixth times 𝑢 to the power of three over two minus one, which we can simplify to give us one-quarter 𝑢 to the power of one-half. To find d𝑢 by d𝑡, we see that 𝑢 is equal to four 𝑡 plus four. And differentiating both sides of this equation, we can see that d𝑢 by d𝑡 is equal to four. So d𝑦 by d𝑡 is equal to one-quarter 𝑢 to the power of a half multiplied by four, which we can simplify to give us 𝑢 to the power of one-half. And then, by substituting 𝑢 is equal to four 𝑡 plus four, we’ve shown that d𝑦 by d𝑡 is equal to four 𝑡 plus four all raised to the power of one-half.

So now that we found d𝑦 by d𝑡 and d𝑥 by d𝑡, let’s clear some space so we can calculate the length of our arc. By using our formula for the length of a curve defined by a pair of parametric equations, we have that 𝑙, the length of the arc we’re asked to calculate, is equal to the integral from zero to two of the square root of 𝑡 squared plus four 𝑡 plus four to the power of a half squared with respect to 𝑡. We can simplify our integrand. 𝑡 all squared is just 𝑡 squared. And one of our laws of exponents tells us that 𝑎 to the power of 𝑏 to the power of 𝑐 is just equal to 𝑎 to the power of 𝑏 times 𝑐. So four 𝑡 plus four raised to the power of a half raised to the power of two is the same as saying four 𝑡 plus four to the power of one-half times two. And one-half times two is just equal to one. So this gives us the length of our arc is equal to the integral from zero to two of the square root of 𝑡 squared plus four 𝑡 plus four with respect to 𝑡.

We might be tempted to try and integrate this by using substitution. However, in this case, there’s a simpler method. We can see that 𝑡 squared plus four 𝑡 plus four is actually equal to 𝑡 plus two all squared. So we can use this to eliminate the square root in our integrand. Using this, we have that the length is equal to the integral from zero to two of the square root of 𝑡 plus two squared with respect to 𝑡. And of course, the square root of 𝑡 plus two all squared is just equal to 𝑡 plus two.

So now we just need to calculate the definite integral from zero to two of 𝑡 plus two with respect to 𝑡. We can do this by using the power rule for integration. We add one to our exponent and divide by this new exponent. This gives us 𝑡 squared over two plus two 𝑡 evaluated at the limits of our integral 𝑡 is equal to zero and 𝑡 is equal to two. Evaluating this at the limits of our integral gives us two squared over two plus two times two minus zero squared over two minus two times zero. We have zero squared over two and two times zero are both equal to zero. Then we see two squared over two is equal to two and two times two is equal to four. So we add these together and we get six.

Therefore, we’ve shown the length of the arc defined by the pair of parametric equations 𝑥 is equal to a half 𝑡 squared and 𝑦 is equal to one-sixth times four 𝑡 plus four to the power of three over two from 𝑡 is equal to zero to 𝑡 is equal to two is six.