Set up an integral that represents the length of the curve.

› Calculus and Analysis › Set up an integral that represents the length of the curve.
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2 years ago
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<= y<= 4
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classicwestconcertdetails Staff answered 2 years ago
Arc Length Formula If f'(y) is continuous on [a,b], then the length of the curve x=f(y), a<= y<= b, is L=int_(a)^(b)sqrt(1+[f'(y)]^(2))dy We have to find length of the curve x=sqrt(y)-y, 1<= y<= 4 f'(y)=(1)/(2sqrt(y))-1 Hence L=int_(1)^(4)sqrt(1+((1)/(2sqrt(t))-1)^(2))dy The problem s us to use a calculator. Using a calculator, i found L=int_(1)^(4)sqrt(1+((1)/(2sqrt(y))-1)^(2))dy~ 3.6095 Result: L=int_(1)^(4)sqrt(1+((1)/(2sqrt(y))-1)^(2))dy~ 3.6095
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