Question

find the radius of curvature of the curve xy=30 at the point (3,10)

Answer 1

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Answer 2

Answer:

Hence, the radius of the curve at the point (3,10) is:

170.699

Step-by-step explanation:

We know that the radius of convergence of any curve y=f(x) at any point M(x,y) is given by:

$R=\dfrac{[1+(y'(x))^2]^{\frac{3}{2}}}{|y"(x)|}$

We have the curve:

$xy=30$

This means that:

$y=\dfrac{30}{x}$

Hence,

$y'(x)=\dfrac{-30}{x^2}$

and

$y"(x)=\dfrac{60}{x^3}$

Hence, on evaluating the expression of R we get the expression as:

$R=\dfrac{(\sqrt{x^4+900})^3}{x^3\times 60}$

Now the value of R at the point (3,10) is:

$R=\dfrac{(\sqrt{3^4+900})^3}{3^3\times 60}$

on solving we get:

$R=170.699$

Hence, the radius of the curve at the point (3,10) is:

170.699