Approximating Nearby Points on a Nonlinear Curve

[anthonym44]

Homework Statement

To the right is the graph of 5x^3y-3xy^2+y^3=6. Verify that (1,2) is a point on the curve. There's a nearby point on the curve whose point is (1.07,u). What is the approx. value for u? There's a nearby point on the curve whose coordinates are (.98,v). What is the approx. value for v? Theres a nearby point on the curve whose point is (w,2.04). What is an approximate value for w?

Homework Equations

f'(a)(x-a)+f(a)
the derivative of the equation is (15x^2-3y^2)/(-5x^3+3x^2y-3y^2)

I know how to do the first part, you plug in the values for x and y and they should = 6, which they do. However, for the remainder of the problem i am completely lost. I'm totally
lost, so if you can offer any input i would appreciate it. Thanks

 

[HallsofIvy]

Homework Statement

To the right is the graph of 5x^3y-3xy^2+y^3=6. Verify that (1,2) is a point on the curve. There's a nearby point on the curve whose point is (1.07,u). What is the approx. value for u? There's a nearby point on the curve whose coordinates are (.98,v). What is the approx. value for v? Theres a nearby point on the curve whose point is (w,2.04). What is an approximate value for w?

Homework Equations

f'(a)(x-a)+f(a)
the derivative of the equation is (15x^2-3y^2)/(-5x^3+3x^2y-3y^2)

I know how to do the first part, you plug in the values for x and y and they should = 6, which they do. However, for the remainder of the problem i am completely lost. I'm totally
lost, so if you can offer any input i would appreciate it. Thanks

The tangent line to a curve, y= f(x), through point (x_0, f(x_0)) on the line is y= f'(x_0)(x- x_0)+ f(x_0) and will give (approximately) the y value for an x value close to x_0. You are told that x_0= 1 and that f(x_0)= 2. What is f'(x_0)? That gives you all the information you need to construct the equation of the tangent line. Then use it with the two new points given to find the missing values.

 

[anthonym44]

Thanks. That helped me, I'm pretty sure I have it now.