Answer to Question #112767 in Differential Geometry | Topology for Akriti

Question

a) Find the gradient of the curve at (2,-3)
b) what does your answer to part (a) tell you about the point (2,-3)

Answer 1

Let us suppose we are given a function $f(x,y,z)$ and we have to find the gradient of this function at a particular point

so gradient can be calculated by $\frac{∂}{∂x}(f(x,y,z))i+\frac{∂}{∂y}(f(x,y,z))j+\frac{∂}{∂z}(f(x,y,z))k$

and after that we will substitute (2,-3) in it and we will get the value of gradient

If the gradient of a vector is not zero, then it will be perpendicular to the tangent line at (2,-3) of the curve that passes through the point, it tells us that the curve changes fastest in the direction of the gradient.