Question Video: Expressing the Length of a Parametric Equation Curve as an Integral | Nagwa Question Video: Expressing the Length of a Parametric Equation Curve as an Integral | Nagwa

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‘₯ = 𝑑² βˆ’ 𝑑 and 𝑦 = 𝑑⁴, where 1 ≀ 𝑑 ≀ 4, as an integral.

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Video Transcript

Express the length of the curve with parametric equations π‘₯ equals 𝑑 squared minus 𝑑 and 𝑦 equals 𝑑 to the fourth power, where 𝑑 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 𝐿 of a curve defined parametrically between the limits of 𝑑 equals 𝛼 and 𝑑 equals 𝛽 is the definite integral between 𝛼 and 𝛽 of the square root of dπ‘₯ by d𝑑 squared plus d𝑦 by d𝑑 squared with respect to 𝑑. Now, our curve is defined parametrically by π‘₯ equals 𝑑 squared minus 𝑑 and 𝑦 equals 𝑑 to the fourth power. And we want to find this arc length between the limits of 𝑑 equals one and 𝑑 equals four. So we let 𝛼 be equal to one, 𝛽 be equal to four. And we see we’re going to need to differentiate π‘₯ and 𝑦 with respect to 𝑑.

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 𝑑 squared is two 𝑑. And when we differentiate negative 𝑑, we get negative one. dπ‘₯ by d𝑑 is, therefore, two 𝑑 minus one. And this satisfies the criteria that the derivative of this function is continuous. d𝑦 by d𝑑 is the first derivative of 𝑑 to the fourth power. That’s four 𝑑 cubed, which is also a continuous function. Now, we noticed that we’re going to have to square these in our formula for the arc length. So let’s work out dπ‘₯ by d𝑑 squared and d𝑦 by d𝑑 squared before substituting into the formula.

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

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