# Question Video: Expressing the Length of a Parametric Equation Curve as an Integral Mathematics • Higher Education

Express the length of the curve with parametric equations π₯ = π‘Β² β π‘ and π¦ = π‘β΄, where 1 β€ π‘ β€ 4, as an integral.

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### Video Transcript

Express the length of the curve with parametric equations π₯ equals π‘ squared minus π‘ and π¦ equals π‘ to the fourth power, where π‘ is greater than or equal to one and less than or equal to four, as an integral.

We recall that the formula for the arc length πΏ of a curve defined parametrically between the limits of π‘ equals πΌ and π‘ equals π½ is the definite integral between πΌ and π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to π‘. Now, our curve is defined parametrically by π₯ equals π‘ squared minus π‘ and π¦ equals π‘ to the fourth power. And we want to find this arc length between the limits of π‘ equals one and π‘ equals four. So we let πΌ be equal to one, π½ be equal to four. And we see weβre going to need to differentiate π₯ and π¦ with respect to π‘.

Now, to differentiate a polynomial term, we simply multiply the entire term by the exponent and then reduce the exponent by one. So the derivative of π‘ squared is two π‘. And when we differentiate negative π‘, we get negative one. dπ₯ by dπ‘ is, therefore, two π‘ minus one. And this satisfies the criteria that the derivative of this function is continuous. dπ¦ by dπ‘ is the first derivative of π‘ to the fourth power. Thatβs four π‘ cubed, which is also a continuous function. Now, we noticed that weβre going to have to square these in our formula for the arc length. So letβs work out dπ₯ by dπ‘ squared and dπ¦ by dπ‘ squared before substituting into the formula.

By distributing the parentheses, we find that two π‘ minus one all squared is four π‘ squared minus four π‘ plus one. And four π‘ cubed all squared is 16π‘ to the sixth power. Our final step is to substitute into the formula for the arc length. And we find that πΏ is equal to the definite integral between one and four of the square root of four π‘ squared minus four π‘ plus one plus 16π‘ to the sixth power dπ‘. We might choose to rewrite the expression inside our root in descending powers of π‘. And when we do, we find that the arc length of the curve defined by our parametric equations for π‘ is greater than or equal to one and less than or equal to four is the integral shown.