Question

Find the radius of curvature of at y= e^ at (0,1)

Answer 1

Step-by-step explanation:

h = x - (dy/dx )[1 + (dy/dx )^2]/ d^2y/dx^2 , at (x,y) = (0,1)

h = 0 - (1) [ 1+ (1)^2] /1 = - 2. A

k = y + [1 + (dy/dx )^2] /( d^2 y / dx^2) = 1. + (1+1)/1 =3 B

Putting A and B together together the centre of curvature is

(h, k ) = C ( - 2 ,3 )

Check ; find the distance from P ( 0 , 1 ) to C ( -2 , 3)

r = √[(- 2 -0)^2 +(3 - 1)^2] = √8 = 2 √ 2 = radius of curvature.

So (h , k) = ( - 2 , 3 ) is the centre.

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Answer 2

Concept

For a given curve, the radius of curvature is the approximate radius of a circle that best fits the curve at a certain point.

Given

y= e^x at (0,1)

Find

The radius of curvature of the given curve at (0,1)

Solution

The radius of curvature for y=f(x) is given by,

$R= \frac{ [1+(\frac{dy}{dx})^2]^\frac{3}{2} } {\frac{d^2x} {d^2y} }$;

The given curve is y = e^x.

Then,

dy/dx = e^x ,

$\frac{d^2y}{dx^2} = e^{x}$,

At the point (0,1)

$\frac{dy}{dx} = e^0 =1$

$\frac{d^2y}{dx^2}= e^0=1$

Radius of curvature

$R= \frac{ [1+(\frac{dy}{dx})^2]^\frac{3}{2} } {\frac{d^2x} {d^2y} }$

R= ([1+(1)^2]^3/2)/1

R=$2^{3/2}$

R= $2\sqrt{2}$

Hence radius of curvature of curve y=e^x at (0,1) is R= $2\sqrt{2}$.

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