1. ## tangent line equation from a gradient vector

Throughout our vector calculus class, we have used
tangent line equation = f(a,b) + f sub x (a,b)(x-a) + f sub y(a,b) (y-b)

On an old final the question is asked to find the equation for the tangent line to a function who's output at (1,3) is = 7 and who's gradient vector (1,3) = 2 i - 5 j

Needless to say since I do not have the function I can not use my nice canned equation to solve this one!

The question is then asked to find the tangent plane to the surface of z = f(x,y) at the point (1,3,7). NO other information is provided?????

I believe in the first question that my gradient vector would be providing how x and y are changing so I should be able to get a slope for my line out of it. I am really not sure how to proceed. Any nudges or pushes toward the solution would be appreciated very much! Frostking

2. Hi

The gradient vector at (1,3) is 2i - 5j
Therefore $\displaystyle \frac{\partial f}{\partial x}(1,3) = 2$ and $\displaystyle \frac{\partial f}{\partial y}(1,3) = -5$