Question

Find the point on the parabola y^2=8x at which the radius of curvature is 125/16
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Answer 1

Answer:

Step-by-step explanation:

We know, rho, radius of curvature =[1+(dy/dx) ^2]^3/2/d^2y/dx^2.......(A)

We are given, radius of curvature=215/16......(a)

and the equation of curve, y^2=8x..... (1)

Differentiating equation... (1) w. r. t x both sides we get,

dy/dx=4/y...(b)

Again differentiating above equation w. r. t x both sides we get,

d^2y/dx^2= -16/y^3...(c)

Using equations (a), (b), (c)... in (A) we get,

125/16=[1+16/y^2]^3/2/-16/y^3

125/16=[y^2+16]^3/2/-16

125/16*16=[y^2+16]^3/2

(125)^2/3=

Answer 2

Given the equation of a parabola $y^2=8x$, Find a point where the radius of curvature is $\frac{125}{16}$.

Explanation:

• The radius of curvature 'R' of any twice differentiable curve $y=f(x)$ at any point $(x,y)$ is given by,    $R=\frac{(1+y'^2)^{\frac{3}{2} }}{|y''|}$  
• Here y' and y'' are the first and second derivatives of the curve at the given point on it.
• Given is the curve $y^2=8x$, let the required point be $(x,y)$.
• Hence, the required derivatives of this curve at the point $(x,y)$ are,
•         $2yy'=8\ \ \ \ \ \ \ \ \ ->y'=\frac{4}{y} \\y''=\frac{-4}{y^2}(y') \ \ \ \ ->y''=\frac{-16}{y^3}$      
• Substituting these values in the formula for the radius of curvature we get,  
•                         $\frac{(1+y'^2)^{\frac{3}{2} }}{|y''|}=\frac{125}{16}\\ ->\frac{(1+(\frac{4}{y})^2)^{\frac{3}{2} }}{|\frac{-16}{y^3}|}=\frac{125}{16} \\->\frac{(y^2+16)^{\frac{3}{2} }}{16}=\frac{125}{16} \\->(y^2+16)^{\frac{3}{2} }= 125=25^{\frac{3}{2} }\\->y^2+16=25\\->y=\pm3 \\->x=\frac{y^2}{8}\\\\ ->y=\pm3, \ x=\frac{9}{8}$    
• Hence, there are two possible points on the parabola for the given radius of curvature $(3,\frac{9}{8} ),\ (-3,\frac{9}{8})$.