If 3ay^2=x(x-a)^2 with a>0, prove that the
radius of curvature at the point (3a, 2a) is
50a/3.
Help with working plz
Just use the formula:
r = (1+y'^2)^(3/2) / |y"|
3ay^2 = x^3-2ax^2+ax
6ay y' = 3x^2-4ax+a
y' = (3x^2-4ax+a)/(6ay)
y" = [(6ay)(6x-4a) - (3x^2-4ax+a)(6ay')]/(36a^2y^2)
Now just plug in (3a,2a) and crank it out. Messy algebra, but not difficult.
To find the radius of curvature at a given point on a curve, we need to use the formula:
R = [(1 + (dy/dx)^2)^(3/2)] / |d^2y/dx^2|
To start, let's find the first derivative dy/dx of the given equation.
Given: 3ay^2 = x(x-a)^2
Differentiating both sides with respect to x using the product rule:
6ay * dy/dx = (x-a)^2 + 2x(x-a) * (1)
Simplifying equation (1):
6ay * dy/dx = (x-a)^2 + 2x(x-a)
Now, let's find the value of y at the given point (3a, 2a) by substituting x = 3a into the given equation:
3ay^2 = (3a)(3a-a)^2
3ay^2 = 9a^2 * a^2
3ay^2 = 9a^4
y^2 = 3a^2
Taking the square root on both sides:
y = √(3a^2) = √(3) * a
Now, let's substitute x = 3a and y = √(3) * a into the equation we obtained after differentiation:
6ay * dy/dx = (x-a)^2 + 2x(x-a)
6a * (√(3) * a) * dy/dx = (3a-a)^2 + 2(3a)(3a-a)
6a^2 * (√(3) * dy/dx) = (2a)^2 + 2(3a)(2a)
6a^2 * (√(3) * dy/dx) = 4a^2 + 12a^2
6a^2 * (√(3) * dy/dx) = 16a^2
(√(3) * dy/dx) = 16/6
dy/dx = 8/(3√(3))
Now, let's find the second derivative d^2y/dx^2 by differentiating dy/dx:
dy/dx = 8/(3√(3))
Differentiating both sides with respect to x:
d^2y/dx^2 = 0
Substituting the values of dy/dx and d^2y/dx^2 into the formula for the radius of curvature:
R = [(1 + (dy/dx)^2)^(3/2)] / |d^2y/dx^2|
R = [(1 + (8/(3√(3)))^2)^(3/2)] / |0|
R = [(1 + (64/(9√(3))))^(3/2)] / 0
Since the denominator is zero, the radius of curvature at the point (3a, 2a) is undefined.
Therefore, the given statement is incorrect.