Question

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places.

x=y²-2y,0≤y≤2

Short answer

The length of the curve is\({\rm{2}}{\rm{.9579}}\).

Step-by-step solution

Arc Length Formula.

If \({\rm{f'(x)}}\)is continuous on\({\rm{(a,b)}}\), then the length of the curve is \({\rm{y = f(x),a}} \le {\rm{x}} \le {\rm{b}}\)

\({\rm{L = }}\int_{\rm{a}}^{\rm{b}} {\sqrt {{\rm{1 + }}{{\left( {{\rm{f'(x)}}} \right)}^{\rm{2}}}{\rm{dx}}} } \)

To determine the curve's length.

Given: \({\rm{x = }}{{\rm{y}}^{\rm{2}}}{\rm{ - 2y,0}} \le {\rm{y}} \le {\rm{2}}\)

\({{\rm{f}}^{\rm{'}}}{\rm{(y) = 2y - 2}}\)

Hence

\({\rm{L = }}\int_{\rm{0}}^{\rm{2}} {\sqrt {{\rm{1 + }}{{\left( {{\rm{2y - 2}}} \right)}^{\rm{2}}}} {\rm{dy}}} \)

Using a calculator to solve a problem. can find

\({\rm{L = }}\int_{\rm{0}}^{\rm{2}} {\sqrt {{\rm{1 + 4}}{{\left( {{\rm{y - 1}}} \right)}^{\rm{2}}}} {\rm{dy}}} \approx {\rm{2}}{\rm{.9579}}\)

Therefore, the length of the curve is\({\rm{2}}{\rm{.9579}}\).