Video Transcript
Find the length of the curve with parametric equations 𝑥 is equal to one plus three 𝑡 squared and 𝑦 is equal to four plus two 𝑡 cubed, where 𝑡 is greater than or equal to zero and 𝑡 is less than or equal to one.
The question wants us to find the length of a curve defined as a pair of parametric equations. And we recall we can calculate the length of a curve defined parametrically as the integral from 𝛼 to 𝛽 of the square root of the derivative of 𝑥 with respect to 𝑡 squared plus the derivative of 𝑦 with respect to 𝑡 squared with respect to 𝑡. And that’s where 𝑡 is greater than or equal to 𝛼 and 𝑡 is less than or equal to 𝛽. So, let’s start by calculating the derivative of 𝑥 with respect to 𝑡 and the derivative of 𝑦 with respect to 𝑡.
We have that 𝑥 is equal to one plus three 𝑡 squared. So, we need to differentiate this with respect to 𝑡. We can do this using the power rule for differentiation. We multiplied by the exponent and then reduced the exponent by one. We have the derivative of one is zero and then the derivative of three 𝑡 squared is six 𝑡. Now, let’s calculate the derivative of 𝑦 with respect to 𝑡. We have that 𝑦 is equal to four plus two 𝑡 cubed. And the derivative of four is just equal to zero and the derivative of two 𝑡 cubed is six 𝑡 squared.
The question wants us to calculate the length of the curve where 𝑡 is between zero and one. So, we’ll set 𝛼 equal to zero and 𝛽 equal to one. Using this, we have the length of our curve 𝐿 is equal to the integral from zero to one of the square root of six 𝑡 all squared plus six 𝑡 squared all squared with respect to 𝑡. We have that six 𝑡 all squared is equal to 36𝑡 squared and six 𝑡 squared all squared is equal to 36𝑡 to the fourth power. We can simplify this by noticing both sums inside of our square root share a factor of 36𝑡 squared. So, this gives us the integral from zero to one of the square root of 36𝑡 squared times one plus 𝑡 squared with respect to 𝑡.
We then see that 36𝑡 squared is a square. It’s equal to six 𝑡 all squared. So, we can evaluate the square root of this. So, the length of our curve is equal to the integral from zero to one of six 𝑡 times the square root of one plus 𝑡 squared with respect to 𝑡. And this is not in a form which we can integrate easily. However, if we notice the derivative of one plus 𝑡 squared is two 𝑡, then we can notice that six 𝑡 is a scalar multiple of two 𝑡. This tells us we can evaluate this integral by using the substitution 𝑢 is equal to one plus 𝑡 squared.
So, starting with the substitution 𝑢 is equal to one plus 𝑡 squared, we differentiate both sides of our equation with respect to 𝑡 to get that d𝑢 by d𝑡 is equal to two 𝑡. And we can use this to get the equivalent statement that d𝑢 is equal to two 𝑡 d𝑡. Now, we want to rewrite the six 𝑡 inside of our integrand as a scalar multiple of two 𝑡. This is three multiplied by two 𝑡. Now, since we’re using integration by substitution on the definite integral, we need to find the new limits of our interval.
We’ll start with the lower limit of our integral. When 𝑡 is equal to zero, we have that 𝑢 is equal to one plus zero squared, which is just equal to one. And for the upper limit of our integral, we have when 𝑡 is equal to one, 𝑢 is equal to one plus one squared, which we can evaluate to give us two. So, by using the substitution 𝑢 is equal to one plus 𝑡 squared, we have that our length is equal to the integral from one to two. We know that two 𝑡 d𝑡 is equal to d𝑢. And we’re using the substitution 𝑢 is equal to one plus 𝑡 squared.
So, we can rewrite the square root of one plus 𝑡 squared as the square root of 𝑢, meaning we need to integrate from one to two three multiplied by the square root of 𝑢 with respect to 𝑢. And we can evaluate this integral using the power rule for integration since the square root of 𝑢 is equal to 𝑢 to the power of one-half. We add one to our exponent and then divide by this new exponent. This gives us two multiplied by 𝑢 to the power of three over two evaluated at the limits of our integral 𝑢 is equal to one and 𝑢 is equal to two.
Evaluating this at the limits of our integral gives us two times two to the power of three over two minus two times one to the power of three over two. We have that one to the power of three over two is just equal to one. And raising a number to the power of three over two is the same as taking the square root and then cubing it. This means that two to the power of three over two is equal to two root two. This means we can simplify this expression to give us four multiplied by the square root of two minus two.
Therefore, we’ve shown the length of the curve with the parametric equations 𝑥 is equal to one plus three 𝑡 squared and 𝑦 is equal to four plus two 𝑡 cubed, where 𝑡 is greater than or equal to zero and 𝑡 is less than or equal to one, is four root two minus two.
