Question

find the radius of curvature for the curve xy3=a4 at (a,a)
​

Answer 1

SOLUTION

TO DETERMINE

The radius of curvature at (a,a) for the curve

$\sf{ x {y}^{3} = {a}^{4} \: }$

FORMULA TO BE IMPLEMENTED

For a curve of the form x = f(y)

The radius of curvature

$\displaystyle \sf{ = \frac{ { \bigg[ \: 1 + { \big( \: \frac{dx}{dy} \big)}^{2} \bigg] }^{ \frac{3}{2} } }{\frac{ {d}^{2} x}{d {y}^{2} }}}$

EVALUATION

Here the given equation of the curve is

$\sf{ x {y}^{3} = {a}^{4} \: }$

$\implies \sf{ x = {a}^{4} {y}^{ - 3}\: }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dx}{dy} = - 3 {a}^{4} {y}^{ - 4} \: }$

Again Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{{d}^{2} x}{d {y}^{2} } = 12 {a}^{4} {y}^{ - 5} \: }$

Now

$\displaystyle \sf{ \frac{dx}{dy} \bigg|_{(a,a)} = - 3 {a}^{4} {a}^{ - 4} = - 3\: }$

$\displaystyle \sf{ \frac{ {d}^{2} x}{d {y}^{2} } \bigg|_{(a,a)} = 12 {a}^{4} {a}^{ - 5} = \frac{12}{a} \: }$

Hence the radius of curvature

$\displaystyle \sf{ = \frac{ { \bigg[ \: 1 + { \big( \: \frac{dx}{dy} \big)}^{2} \bigg] }^{ \frac{3}{2} } }{\frac{ {d}^{2} x}{d {y}^{2} }}}$

$\displaystyle \sf{ = \frac{ { \bigg[ \: 1 + { \big( - 3 \big)}^{2} \bigg] }^{ \frac{3}{2} } }{ \frac{12}{a} }}$

$\displaystyle \sf{ = \frac{ { \bigg[ \: 1 + 9 \bigg] }^{ \frac{3}{2} } }{ \frac{12}{a} }}$

$\displaystyle \sf{ = \frac{a}{12}{ \big(1 + 10 \big)}^{ \frac{3}{2} } \: }$

$\displaystyle \sf{ = \frac{a}{12}10 \sqrt{10} }$

$\displaystyle \sf{ = \frac{5a \sqrt{10} }{6}}$

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

The length of the latus rectum of the parabola

13[(x-3)^2+(y-4)^2 )= (2x-3y+ 5)^2 is

https://brainly.in/question/24980361