Question

can i get some help? i'm a little lost.

Suppose that you are climbing a hill whose shape is given by z=833−0.05x^2−0.02y^2, and that you are at the point (90, 80, 300).
In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest?
⟨ , ⟩
If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?

Answer 1

i thought it would just be gradient = <-.1x, -.04y> making part a be <9,3.2>, but that wasn't right.

Answer 2

gradient = <-.1x, -.04y>
at x = 90, y = 80:
gradient = <-9, -3.2>
"In which direction (UNIT VECTOR) should you proceed ...."
| gradient | = sqrt( (-9)^2 + (-3.2)^2 ) = ?
unit vector = (-9i - 3.2j) / | gradient | = ?

Answer 3

I am getting -0.9422i - 0.3350j

Answer 4

oh. i never know when i have to divide by the magnitude and when i don't.....

Answer 5

If they ask for UNIT VECTOR you will have to divide by the magnitude.
Only then the magnitude of the vector will be 1.

Answer 6

so how do i find the degree? part b

Answer 7

i know Duf= gradf *u *costheta, but not sure what to use in that equation.

Answer 8

The initial position vector is <90, 80, 300>. Find the magnitude.
The gradient vector is <-9, -3.2, 0>. Find the magnitude.
cos(theta) = | gradient | / | position |
set calculator to radian mode and take the inverse since they want the angle in radians.

Answer 9

put this into webwork....arccos(sqrt((-9)^2+(-3.2)^2)/sqrt(90^2+80^2+300^2)) and it says incorrect....

Answer 10

Is the mode set to radians?

Answer 11

in webwork, you can't change degrees or radians. it always does it in radians.

Answer 12

I am getting 1.5412 radians. Can you try that?

Answer 13

1.54124358964091

Answer 14

Is it saying that is wrong?
Did it accept answer to part a) ?

Answer 15

getting a screenshot. two secs.

Answer 16

ok, i'll check with the teacher. thanks

Answer 17

How many tries are you allowed?

Answer 18

unlimited

Answer 19

Wondering if sticking a pi - in front would work.

Answer 20

pi minus arccos(....)

Answer 21

nope

Answer 22

Alright I'll leave it to your teacher.

Answer 23

thanks!

Answer 24

you are welcome. If you do find out the answer you can post it here and I'll check it the next time.

Answer 25

The gradient vector is <-9, -3.2>
@saiken2009 Could you try arctan( (-9)^2 + (-3.2)^2 ) and tell me if it accepts the answer?

Answer 26

no, it doesn't.