To find the directional derivative calculate the gradient for the function h and than take the dot product of the gradient for the function h with the unit vector in the direction of v.

Hint: v is not a unit vector so you will have to find the unit vector in the direction of v using the following formula

Uv = v*(1/ |v|)

where |v| is the length or magnitude of the vector v.

Uv: stands for the unit vector in the direction of v.

For the gradient of h I got

grad h = – e^(-x^2-y^2), – e^(-x^2 – y^2)

check this answer. Once you have the expression for the gradient evaluate it at the point (90, 0). Then take the dot product of the gradient for h with the unit vector you find for the vector v and you should have the directional derivative.

Remember the dot product should give a scalar as a result.

B

For the iterated integral, evalute it as you would normally.

In this case you have to evaluate with respect to x (with the dx term) first since the limits of integration are variable (in terms of y) instead of a number. When you integrate with respect to x you treat the y terms as constants.

After you evaluate from y to 2y the whole expression should be in terms of y and then you are left with a single variable integral and you integrate in terms of y from 0 to 2.

C

For this problem Draw the region you are integrating (which is in R2, the xy-plane). Draw the lines y= x and y=2 which will give you a triangular region. This will help you determine your limits of integration. You are integrating the function 4-y^2. Since you are doing a double integral it should be of the form

SS 4-y^2 dx dy or SS 4-y^2 dy dx

depending on how you choose the limits of integration (Type 1 or Type 2 region)