(a) This vector is in the direction of -∇f(2,6).

(The negative sign is for decreasing most rapidly.)

∇f = <(e^x - 5) sin y, (e^x - 5x) cos y>

==> ∇f(2, 6) = <(e^2 - 5) sin 6, (e^2 - 10) cos 6>.

So, the desired direction is -∇f(2, 6) = <-(e^2 - 5) sin 6, -(e^2 - 10) cos 6>.

Note: If you need this to be a unit vector, divide each entry by ||∇(2,6)||.

----

(b) First of all the equation of the tangent plane to f at (2, 6) is given by

z - f(2, 6) = (e^2 - 5) sin 6 · (x - 2) + (e^2 - 10) cos 6 · (y - 6).

==> (e^2 - 5) sin 6 · (x - 2) + (e^2 - 10) cos 6 · (y - 6) - 1(z - (e^2 - 10) sin 6)) = 0.

Hence, <(e^2 - 5) sin 6 , (e^2 - 10) cos 6, -1> is a normal vector (coming from the coefficients of x, y, and z) to the tangent plane at the given point.

Since two vectors are perpendicular if their dot product is zero, we need

<(e^2 - 5) sin 6 , (e^2 - 10) cos 6, -1> · <3, 2, a> = 0

==> 3(e^2 - 5) sin 6 + 2(e^2 - 10) cos 6 - a = 0

==> a = 3(e^2 - 5) sin 6 + 2(e^2 - 10) cos 6.

I hope this helps!

# Consider the function f(x,y)=(e^x-5x)sin(y).?

Suppose S is the surface z=f(x,y).

(a) Find a vector which is perpendicular to the level curve of f through the point (2,6) in the direction in which f decreases most rapidly.

(b) Suppose v= 3i+2j+ak is a vector in 3-space which is tangent to the surface S at the point P lying on the surface above (2,6). What is a?

(a) Find a vector which is perpendicular to the level curve of f through the point (2,6) in the direction in which f decreases most rapidly.

(b) Suppose v= 3i+2j+ak is a vector in 3-space which is tangent to the surface S at the point P lying on the surface above (2,6). What is a?

## 1 Answers

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