Question

Thanks

Answer 1

Directional Derivative of `f` in the direction of `u` is
the rate at which `f` changes in the direction of `u`

and is given by
\(\Large \nabla_uf = \nabla f \times \dfrac{u}{|u|}\)

Answer 2

so first form a vector 'u'
then find \(\nabla f\)
and use the above formula!

Answer 3

Thank you,
so for my example Df = exp(-x^2 - y^2) * (-2x -2y) correct?

and u = (1/sqrt(5) , 2 / sqrt(5) ) ?

Answer 4

if by total derivative, you meant \(\nabla f\)
then you were correct :)

Answer 5

u is correct, let me check for Df

Answer 6

wouldn't Df be a vector ?

Answer 7

so Df = (Dx , Dy) ?

Answer 8

yesss

Answer 9

ah, would make my workings much simpler, currently its a mess :)

Answer 10

so (-2xexp(-x^2 - y^2) , -2yexp(-x^2 -y^2) ) ?

Answer 11

that is correct :)

Answer 12

so the final result would be just u multiplied by Df and then just input the values of (1 , 2) for x and y ?

Answer 13

and sorry about the notation, i meant
\(\Large \nabla_uf = \nabla f . \dfrac{u}{|u|}\)

the dot product between \(\nabla f\) and unit vector u

Answer 14

For calculating value of Dx would I take y = 0 and x = 1 or y = 2 and x = 1?

Answer 15

you DD will be
\(\large ( (-2xe^{(-x^2 - y^2)} , -2ye^{(-x^2 -y^2)} ) ).(1/\sqrt5 , 2 / \sqrt5 )\)

Answer 16

don't plug in anything for x and y
just calculate the dot product

Answer 17

oh really? wow okay.. thanks seems simple enough, but why don't you plug in values for Df ?

Answer 18

oh yeah, you plug in x=1,y=2

Answer 19

sorry, bit rusty on this

Answer 20

because you get
\(\Large \nabla f|_{(x_0,y_0)}\)

Answer 21

okay so Df(1,2) = (-2exp(-3) , -4exp(-3) ) ?

Answer 22

you mean
Df(1,2) = (-2exp(-5) , -4exp(-5) )
??

Answer 23

yes sorry my bad, i did (-1)^2

Answer 24

go ahead

Answer 25

why do you want to plugin a vector into a scalar function ?

Answer 26

directional derivative gives you the deriivative in the direction (1,2) as function of x and y
it need not be constant

Answer 27

ok, so just
\(\large ( (-2xe^{(-x^2 - y^2)} , -2ye^{(-x^2 -y^2)} ) ).(1/\sqrt5 , 2 / \sqrt5 )\)

Answer 28

So the final result is
\[(\frac{ -2 }{ \sqrt{5}}e^{-5} , \frac{ -8 }{ \sqrt{5}}e ^{-5})\]

Answer 29

http://gyazo.com/0990d44103fe011199509509d6a52dc8

Answer 30

oh so I dont do Df(1 ,2)

Answer 31

in any case, final answer won't be a vector

Answer 32

notice the dot product

Answer 33

can you tell me where u got that idea of plugging (1,2) into the derivatives ?

Answer 34

I assumed that it wanted to find it in terms for x = 1 y = 2 but clearly not..

Answer 35

take a look at the attached graph

Answer 36

do you see a plane in cyan color and a surface in red color ?

Answer 37

yes

Answer 38

directional derivative along the vector (1,2) gives you the slope of tangent to that plane-surface intersection curve

Answer 39

just find the dot product and you're done :

\[\large ( (-2xe^{(-x^2 - y^2)} , -2ye^{(-x^2 -y^2)} ) ) \bullet (1/\sqrt5 , 2 / \sqrt5 )\]

Answer 40

I dint see the graph ;-}

Answer 41

Okay thanks I will reply with my result soon :)

Answer 42

are u in university right now @Marki ?

Answer 43

see if u can open that file

Answer 44

\[e ^{-x^{2} - y^{2}} (-\frac{ 2x }{ \sqrt{5} } - \frac{ 4y }{ \sqrt{5} }\]

Answer 45

) at the end

Answer 46

looks good !

Answer 47

:D thank you! I have two more questions relating to f(x,y) = exp(x^2 - y^2) if you are able to help?

Answer 48

wil try, ask..

Answer 49

okay so how do I find the contour lines of f(x,y) and sketch the graph?
Im assuming that the contour lines will be circular? because of x^2 and y^2

Answer 50

contour curves are formed by intersecting a horizontal plane with the surface of f(x,y)

Answer 51

\[ z = f(x,y) = \exp(-x^2 -y^2)\]
\[ z = k\]

Answer 52

\[\exp(-x^2-y^2) = k\]

Answer 53

\[-x^2-y^2 = \ln k\]

Answer 54

\[x^2+y^2 = -\ln k\]

Answer 55

so you do get circles as your contour curves
but notice that there will be restriction on the level because you want \(\ln k\) to be negative to stay in real plane

Answer 56

so I should just take a point say (0,0) then sketch the graph? with the contour curves, but how do you know the points of the curves, honestly this is all confusing :S

Answer 57

\[x^2+y^2 = -\ln k\]

you get a contour curve by giving a value to \(k\)
clearly they all are circles whose center is origin, yes ?

Answer 58

and radius = \(\sqrt{-\ln k }\)

Answer 59

sketch them should be trivial
just assign specific values to \(k\) and sketch them

Answer 60

yes I understand that part, because of x^2 + y^2 it will produce circles, and that their center will be 0 because its just a standard point, but what will the actual graph look like? and how many contour curves should I produce?

Answer 61

as many as you wish

Answer 62

4-5 curves will do to get some idea aboutt he the 3 dimensional surface

Answer 63

what sort of scale and intervals? start with k = 1 increment by 1 till k = 6?

Answer 64

first find out the restriction on k

Answer 65

your equation of contour curves :
\[x^2+y^2 = -\ln k\]

Answer 66

radius of circle makes sense only when the right hand side is a positive number yes ?

Answer 67

k > 0

Answer 68

wait
k < 0

Answer 69

logarithm is not defined for negative numbers.

Answer 70

you cant make the right hand side positive..

Answer 71

you can, work it

Answer 72

0 > k < 1

Answer 73

Yep!

Answer 74

0 < k < 1 sorry

Answer 75

yes looks good, pick 4-5 values for k in that range and plot ur circles

Answer 76

i chose k = 0.1 k = 0.3 k = 0.5 k =0.7 and k = 0.9
I got a range of R smallest at 0.32 and largest at 1.52 sound okay?

Answer 77

looks good, plot few more close to 0.1 if you want :

try below values for k:
```
0.01
0.05
0.1
0.3
0.5
0.9
```

Answer 78

okay thanks I have a rough sketch now, how do I go about sketching the graph? I tried wolfram alpha and sort of like a mountain..

Answer 79

whats the highest value of level ? (z or k)

Answer 80

0.9 for my contours

Answer 81

yes put ur contour curves along z axis with z = k