Let 2x^2 y - 3x^3 = 5y^3 - 4x + 13?

Question

a) Find an expression for dy/dx

b) Find the gradient of the tangent to the curve at the point
(-2,-1).

c) Find the equation of the normal to the curve at the point
(-2,-1)

Please answer all 3 qs its urgent

Answer 1

a. Differentiate with respect to x.

4xy + 2x^2 dy/dx - 9x^2 = 15y^2 dy/dx - 4
4xy + (2x^2 - 15y^2) dy/dx = -4
dy/dx = -(4 + xy)/(2x^2 - 15y^2)

This is all you need since you have values of x and y.

b. Evaluate dy/dx at the point (-2,-1) to get the gradient
dy/dx = -(4 + (-2)(-2))/(2(-2)^2 - 15(-1)^2)
dy/dx = -8/(8 - 15) = 8/7

c. For the normal: (slope of normal) = -1/m = -7/8
Using the slope intercept form:
y = (-7/8)x + b
Since the line goes through the point (-2,-1):
b = -1 +(7/8)(-2) = -11/4

So the normal line is: y = (-7/8)x - 11/4

Answer 2

sorry...i dont know..