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Mathematics 9 Online
OpenStudy (anonymous):

Consider the function f(x,y)=3x2+3y2. Find the the directional derivative of f at the point (1,−1) in the direction given by the angle θ=2π3. Find the unit vector which describes the direction in which f is increasing most rapidly at (1,−1).

OpenStudy (anonymous):

i got the first part as 6cos(2pi/3)-6sin(2pi/3), but shouldn't part b just be the gradient of f at (1,-1)?

OpenStudy (amistre64):

the gradient in a given direction

OpenStudy (amistre64):

|dw:1412709752318:dw|

OpenStudy (amistre64):

if we take a plane and slice the surface we create a curve in a plane, a plane the is directed at a given angle. the slope at a given point depends in the direction we have heading

OpenStudy (anonymous):

would it be in the same direction as the first u? cos 2pi/3 and sin2pi/3?

OpenStudy (amistre64):

reviewing some stuff to remember what i forgot http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/grad/grad.html

OpenStudy (amistre64):

gradient dotted with the direction vector

OpenStudy (amistre64):

u = cos(2pi/3) , sin(2pi/3)

OpenStudy (amistre64):

3x2+3y2 gradf = 6x, 6y

OpenStudy (amistre64):

Duf = (cos(2pi/3) 6x , sin(2pi/3) 6y)

OpenStudy (anonymous):

yes, that is the answer to part a. that's exactly what i got for part a.

OpenStudy (amistre64):

so part b is: Find the unit vector which describes the direction in which f is increasing most rapidly at (1,−1).

OpenStudy (amistre64):

refresh to get rid of the <?> marks

OpenStudy (amistre64):

the gradient has a property that it actually points in the direction of fastest growth ... but i cnat figure out why

OpenStudy (anonymous):

i think i figured it out.....it's the gradient over the magnitude of the gradient....

OpenStudy (amistre64):

"the gradient vector points in the direction of greatest rate of increase of f(x,y)" 6x, 6y at (1,-1) is 6,-6 unit is yes, dividing a vector by its length

OpenStudy (anonymous):

thanks. i got the right answer.

OpenStudy (amistre64):

youre welcome

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