Question

Find the radious of the curvature of the curve xy =30at the point (3,10)

Answer 1

Answer:

10. 11. 12. 13. 14. Radius of curvature:

Step-by-step explanation:

Explanation (g) xy = coat (c,c) (h) y' = x* + 8 at (–2,0) (i) x = 3a cos 0– a cos 30, ..

Answer 2

To find:

The radius of curvature of the curve $\displaystyle xy=30$ at the point $\displaystyle (3,10)$

Step-by-step explanation:

The given curve is $\displaystyle xy=30$

$\displaystyle \Rightarrow y=\dfrac{30}{x}$, where we must consider $\displaystyle x$ being a non-zero value

Now, differentiating both sides with respect to $\displaystyle x$, we get

$\displaystyle \quad\dfrac{dy}{dx}=30\dfrac{d}{dx}(\dfrac{1}{x})$

$\displaystyle \Rightarrow \dfrac{dy}{dx}=-\dfrac{30}{x^{2}}$

Again, differentiating botb sides with respect to $\displaystyle x$, we get

$\displaystyle \quad\dfrac{d}{dx}(\dfrac{dy}{dx})=-30\dfrac{d}{dx}(\dfrac{1}{x^{2}})$

$\displaystyle \Rightarrow \dfrac{d^{2}y}{dx^{2}}=\dfrac{60}{x^{3}}$

So, we have:

• $\displaystyle y_{1}=-\dfrac{30}{x^{2}}$

• $\displaystyle y_{2}=\dfrac{60}{x^{3}}\neq 0$

Let us put the values in the following formula:

$\displaystyle \quad \rho=\dfrac{(1+y_{1}^{2})^{3/2}}{y_{2}}$

$\displaystyle \Rightarrow \rho=\dfrac{(1+\dfrac{900}{x^{4}})^{3/2}}{\dfrac{60}{x^{3}}}$

$\displaystyle \Rightarrow \rho_{x=3}=\dfrac{(1+\dfrac{900}{81})^{3/2}}{\dfrac{60}{27}}$

$\displaystyle \Rightarrow \rho_{x=3}=18.97$

Answer:

The radius of curvature of the curve $\displaystyle xy=30$ at the point $\displaystyle (3,10)$ is 18.97.

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