Question

Compute the directional derivative of the function f(x,y) = 4xy + 3y^2 at the point P(3,-4) in the direction of the vector <-3,-1>.

Answer 1

fisrt find the gradient of\[f(x,y)=4xy+3y^2\]

Answer 2

can u do it?

Answer 3

No. It's new to me, can't figure out the right way to think of it.

Answer 4

but you need to lnow how to find gradient, without this u cant solve it

Answer 5

ok i can solve it for u ,do u know partial derivatives?

Answer 6

Not really getting the hang of it. Is it this?
x partial = 4y
y partial = 4x + 6y

Answer 7

yes so the gradient will be
\[\nabla f(x)=4y \hat{i}+(4x+6y) \hat{j}\]

Answer 8

so, plug in -3 and -1 for i and j?

Answer 9

no now plug 3,-4 for x and y

Answer 10

what u get ?

Answer 11

\[\nabla f(x)=-14 \hat{i}+12 \hat{j}\]

Answer 12

right?

Answer 13

Yes. How do you enter the gradient symbol?

Answer 14

just type "nabla" in the equation editor

Answer 15

now the vector given is\[-3 \hat{i}- \hat{j}\]
right?

Answer 16

Yes.

Answer 17

so now its unit vector will be--\[U_v=\frac{ -3 \hat{i}-\hat{j} }{ \sqrt{3^2+1^2} }=\frac{ -3 \hat{i}-\hat{j} }{ \sqrt{10} }\]
right?

Answer 18

ok?

Answer 19

how?

Answer 20

take a vector V=ai+jb then its unit vector is\[U_v=\frac{ the~vector }{ It's~magnitude }=\frac{ a \hat{i}+b \hat{j} }{ \sqrt{a^2+b^2} }\]

Answer 21

ok

Answer 22

now finnaly directional derivative is\[\large D.D=unit~vector*gradient_{at~the~given~point}\\=\frac{ -3 \hat{i}-\hat{j} }{ \sqrt{10} }*(-14 \hat{i}-12 \hat{j})=\frac{ 42+12 }{ \sqrt{10} }=\frac{ 54 }{ \sqrt{10} }\]

Answer 23

ok?

Answer 24

hmm.. Answer is not correct

Answer 25

hey ahh we did a small mistake

Answer 26

12 was supposed to be positive?

Answer 27

\[[\nabla f(x)]_{(3,-4)}=-16 \hat{i}-12 \hat{j}\]

Answer 28

whats the answer
\[60/\sqrt{10}\]

Answer 29

Ah. Yes, that's right. Thank you.

Answer 30

yw!!