Question

find the centre of curvature at the point (1,-1)of the curve y=x³-6x²+3x+1 hence find the equaton of the circle of curvature at this point​

Answer 1

Answer: please find the attached file.

Step-by-step explanation:

Answer 2

Answer:  center of the circle of  y = x³- 6x² + 3x + 1 at (1, -1) is equal to (-36, -7.5)

The equation of the circle of curvature = $(x+36)^{2} + (y+7.5)^{2} = 1407$

Step-by-step explanation:

We know that the formula for the center of curvature of a given curve at (a, b) = (X, Y),

Where,

X = $a - \frac{(1+y_{1} ^{2})y_{1} }{y_{2} }$ -------------------------(a)

Y = $b + \frac{(1+y_{1} ^{2}) }{y_{2} }$     -------------------------(b)

y₁ = First derivative of the given equation

y₂ = Second derivative of the given equation

Given equation : y = x³- 6x² + 3x + 1--------------------(1)

y₁ = 3x² - 12x + 3

y₁(1,-1) = -6

y₂ = 6x - 12

y₂(1,-1) = -6

Putting the values of y₁ and y₂ in eq (a) and (b) we get

X = $1 - \frac{(1+(-6 ^{2}))(-6) }{(-6) }$   = -36

Y = $-1 + \frac{(1+(-6 ^{2}) ) }{-6 }$      = -7.5

So, the center of the circle of eq (1) is equal to (-36, -7.5)

Now, the radius of curvature:  R = $\frac{(1+y_1^2)^{3/2}}{y_2}$ = $\frac{(1+(-6)^2)^{3/2}}{-6}$

R² = 1407

And the equation of the circle of curvature = $(x-X)^{2} + (y-Y)^{2} = R^2$

= $(x+36)^{2} + (y+7.5)^{2} = 1407$