a curve has equation y= 1/x^2 + 4x

a) find dy/dx

b) the point P(-1,-3) lies on the curve. Find an equation of the normal to the curve at the point P

c) find an equation of the tangent to the curve that is parallel to the line y= -12x

Guest Nov 21, 2018

#1**+2 **

Write the equation as y = x^{-2} + 4x.

a) Then dy/dx = -2x^{-3} + 4

b) The slope of the *tangent* line at P is thus (dy/dx)_{tangent} = -2(-1)^{-3} + 4

The slope of the *normal* line at P is given by (dy/dx)_{normal} = -1/(dy/dx)_{tangent} .

c) The tangent line parallel to y = -12x will have slope -12, so will be of the form y = -12x + k, where k is a constant.

We must have -2x^{-3} + 4 = -12 for the slopes to match, which gives us the value of x at which this tangent line touches the curve. Plug this value into y = x^{-2} + 4x to find the corresponding value of y. Plug these values of x and y into y = -12x + k and rearrange to find k.

I'll leave you to do the actual calculations!

Alan
Nov 21, 2018