Regard as the independent variable and as the dependent variable and find the slope of the tangent line to the curve (x^2 + y^2)^2 - 4x^2 y = 225 at the point (x,y) = (4,1)
Thank you very much for helping me.
A little implicit differentiation helps out here.
2(x^2 + y^2)(2x + 2yy') - (8xy + 4x^2 y') = 0
4x(x^2 + y^2) + 4y(x^2 + y^2) y' = 8xy + 4x^2 y'
y'(4x(x^2 + y^2) - 4x^2) = 8xy - 4x(x^2 + y^2)
y' = 4x(2y - (x^2 + y^2))/(4x^3 - 4x^2 + 4xy^2)
at (4,1) we have
y' = 4*4(2-17)/(256 - 64 + 16)
y' = -16*15/208 = -1.15
Better double-check the algebra.
To find the slope of the tangent line to the curve, we need to find the derivative of the equation with respect to x.
The given equation is (x^2 + y^2)^2 - 4x^2y = 225.
Let's differentiate both sides of the equation implicitly with respect to x:
2(x^2 + y^2)(2x + 2y(dy/dx)) - 8xy - 4x^2(dy/dx) = 0
Now, we need to find the value of dy/dx at the point (x,y) = (4,1).
Substituting the values into the equation, we have:
2(4^2 + 1^2)(2(4) + 2(1)(dy/dx)) - 8(4)(1) - 4(4^2)(dy/dx) = 0
Simplifying further:
2(16 + 1)(8 + 2(dy/dx)) - 32 - 64(dy/dx) = 0
2(17)(8 + 2(dy/dx)) - 32 - 64(dy/dx) = 0
34(8 + 2(dy/dx)) - 32 - 64(dy/dx) = 0
272 + 68(dy/dx) - 32 - 64(dy/dx) = 0
204 - (4(dy/dx)) = 0
4(dy/dx) = 204
dy/dx = 204/4
dy/dx = 51
So, the slope of the tangent line to the curve at the point (x,y) = (4,1) is 51.