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2022-08-19

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 = \sqrt{y} - y, 1 \leq y \leq 4

Answer & Explanation

Rachael Trevino

Beginner2022-08-20Added 9 answers

Arc Length Formula
If f'(y) is continuous on [a,b], then the length of the curve $x = f (y), a \leq y \leq b$ , is
$L = \int_{a}^{b} \sqrt{1 + {[f^{'} (y)]}^{2}} d y$
The length of the curve $x = \sqrt{y} - y, 1 \leq y \leq 4$ must be determined.
$f^{'} (y) = \frac{1}{2 \sqrt{y}} - 1$
Thus
$L = \int_{1}^{4} \sqrt{1 + {(\frac{1}{2 \sqrt{t}} - 1)}^{2}} d y$
The issues require the use of a calculator.
Using a calculator, i found
$L = \int_{1}^{4} \sqrt{1 + {(\frac{1}{2 \sqrt{y}} - 1)}^{2}} d y \approx 3.6095$
Answer:
$L = \int_{1}^{4} \sqrt{1 + {(\frac{1}{2 \sqrt{y}} - 1)}^{2}} d y \approx 3.6095$

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