Question

let f(x,y) = square-root(1+4x+y^2) and let P be the point (1,2)

a) at P what is the direction of maximal increase for the function f? give your answer as a init vector.

Answer 1

what do you know about gradient vector ?

Answer 2

You want to maximize \(df\).

Answer 3

I know that is \[f_x i +f_y j + f_z z\]

Answer 4

Get \(df\) using the chain rule.

Answer 5

the integral of f(x,y)?

Answer 6

No, the derivative

Answer 7

derivative to respect to y or x?

Answer 8

griadent points in the maximum increasing direction of the function
so you just need to find the gradient at given point

Answer 9

the given function is in two variables, so gradient would be just
\[\langle f_x, ~f_y\rangle \]

Answer 10

do i have to sub in the point P?

Answer 11

yes you want gradient at the given point

Answer 12

gradient is a vector field so you will be having one vector at each point

Answer 13

\[gradf = 2/(\sqrt{y^2+4y+1}+y/\sqrt{4x+y^2+1}\]

Answer 14

thats not a vector ^

Answer 15

\[grad f = \frac{ 2 }{ \sqrt{y^2+4y+1}} i + \frac{ y }{ \sqrt{4x+y^2+1} } j\]

Answer 16

looks better :)

Answer 17

did i get the gradient right?

Answer 18

looks you have few typoes

Answer 19

where?

Answer 20

in the denominator expression

Answer 21

\[grad f = \frac{ 2 }{ \sqrt{1+4x+y^2}} i + \frac{ y }{ \sqrt{1+4x+y^2} } j \]

right ?

Answer 22

oh ok I see where I made a mistake

Answer 23

so I have to sub in the point (1,2)

Answer 24

yep that gives you gradient vector at that particular point

Answer 25

\[\frac{ 2 }{ \sqrt{9} } i + \frac{ 2 }{ \sqrt{9} } j \]

Answer 26

looks good ! read the question again, they want unit vector so divide by the magnitude

Answer 27

divide by the magnitude of the gradf?

Answer 28

hats the unit vector in the direction of
\[\frac{ 2 }{ \sqrt{9} } i + \frac{ 2 }{ \sqrt{9} } j\]

?

Answer 29

\[
df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = \left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right\rangle \cdot d\langle x,y\rangle
\]We can always let \( d\langle x,y\rangle\) be parallel to \( \left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right\rangle \) to maximize \(df\).

Answer 30

yeah we are maximizing \[\nabla f \cdot \dfrac{d\vec{r}}{dt} \]

Answer 31

\[\frac{ 2 }{ \sqrt{9} } i + \frac{ 2 }{ \sqrt{9} } j\]

Notice this points in the same direction as

\[1i + 1j\]

Answer 32

so that could be the unit vector?

Answer 33

i + j

Answer 34

i+j is NOT an unit vector

Answer 35

the magnitude of unit vector has to be 1

Answer 36

whats the magnitude of vector i+j ?

Answer 37

2

Answer 38

magnitude of ai+bj is the distance from origin to point (a, b)

Answer 39

squareroot (2)

Answer 40

so whats the unit vector in the direction of i+j ?

Answer 41

squareroot(1^2 +1^2)

Answer 42

the unit vector in the direction of i+j would be
\[\dfrac{1}{\sqrt{1^2+1^2}}i + \dfrac{1}{\sqrt{1^2+1^2}} j\]