Need your help with this problem please....

[samiam]

Could somebody please help me solve part b?

Suppose f is differentiable and grad f (-2,5) = -3i + 4j (i and j are vectors), and f (-2,5) = 1.

a) Give the equation of the tangent plane to the graph of f at x = -2, y = 5.


answer :

m = -3 (slope of x) and n = 4 (slope of y)

so equation of the tangent plane to f(-2,5) = -3(x+2) + 4(y-5) + 0(z-0)
= -3x + 4y - 26 = 0


b) Give the equation of the tangent line to the contour for f at x = -2, y = 5.


?????? (contour for f at x = -2, y = 5 is 1 I believe)

 

[doug reich]

Wouldn't the vector tangent to the contour be the gradient? So the contour is

x = -3t - 2
y = 4t + 5
z = 1

or, equivalently

( \vec T = \vec{ \Del f } t + \vec x_0)

(can't seem to find the gradient latex, so Del will have to do)

 

[samiam]

Doug thanks.

That seems like something beyond Calc 3 which this is. Or at least so far.

The gradient vector points in the direction of increasing f at (a,b) and is perpendicular to the contour of f through (a,b).

The question simply asks for the equation of the line tangent to the contour at 1. It's a simple x and y equation of a line.

Since we have the gradient vector and it is at a 90 degree angle to the contour of 1 at (-2,5) am I supposed to take this gradient vector and make it at a zero degree angle to the contour.

How am I supposed to this if this is the right approach?

Sam