OpenStudy (anonymous):

consider f(x,y) and the point P=(1,2). The directional derivatives of f at P along the unit vector u=I is 12 and along the unit vector v=1/sqr root(3){i+j+k} is 14. Determine the vaule of directional derivatives of f along the unit vector w=1/2*sqr root(5){3i+4j+5k}

5 years ago
OpenStudy (turingtest):

what is vector u ?

5 years ago
OpenStudy (anonymous):

\[v=1/\sqrt{3}(i+j+k) \]

5 years ago
OpenStudy (anonymous):

w=1/2sqrt{5}(3i+4j+5k)

5 years ago
OpenStudy (anonymous):

\[w=1/2\sqrt{5}(3i+4j+5k)\]

5 years ago
OpenStudy (anonymous):

u is a unit vector u=I

5 years ago
OpenStudy (anonymous):

the problem is function to be defined from two directional derivatives

5 years ago
OpenStudy (anonymous):

directional derivative of f(x,y) in direction of v is defined as: \[\nabla f . v\] so: \[\nabla f . u = 12 \] \[\nabla f . v = 14 \] let nabla f be (x,y,z) just need 1 more equation to be able to solve this to get nabla f. Still thinking...

5 years ago
OpenStudy (anonymous):

yes myko the main part to identify the function f(x,y)

5 years ago
OpenStudy (turingtest):

u is which unit vector? does it not give any more info?

5 years ago
OpenStudy (anonymous):

for u the directional derivatives is 12

5 years ago
OpenStudy (turingtest):

so all the info you have is\[\|\vec u\|=1\]\[\nabla f\cdot\vec u=12\]\[\nabla f\cdot\vec v=\frac1{\sqrt3}\nabla f\cdot\langle1,1,1\rangle=14\]\[\nabla f\cdot\vec w=\frac1{2\sqrt5}\nabla f\cdot\langle3,4,5\rangle\]correct?

5 years ago
OpenStudy (anonymous):

yes and point P=(1,2) and u=I (capital letter i)

5 years ago
OpenStudy (turingtest):

what does capital i mean?, \[\vec u=\langle1,0,0\rangle\]

5 years ago
OpenStudy (anonymous):

yes u=(1,0,0)

5 years ago
OpenStudy (turingtest):

so then...\[\nabla f(1,2)\cdot\vec u=f_x(1,2)=12\]\[\nabla f(1,2)\cdot\vec v=\frac1{\sqrt3}f_x(1,2)+\frac1{\sqrt3}f_y(1,2)=\frac1{\sqrt3}[12+f_y(1,2)]=14\]\[f_y(1,2)=14\sqrt3-12\]can you solve it now?

5 years ago
OpenStudy (anonymous):

i will use the vector w and provide the the value of Fx and Fy ?

5 years ago
OpenStudy (turingtest):

yep

5 years ago
OpenStudy (anonymous):

let me try give me moment

5 years ago
OpenStudy (anonymous):

the answer i got is\[(42 \sqrt{3} - 5)/ \sqrt{5}\] am i right ?

5 years ago
OpenStudy (turingtest):

I got\[\nabla f\cdot\vec w=\frac1{2\sqrt5}\langle12,14\sqrt3-12,0\rangle\cdot\langle3,4,5\rangle=\frac1{2\sqrt5}(36+56\sqrt3-48+0)\]\[=\frac{28\sqrt3-6}{\sqrt5}\]

5 years ago
OpenStudy (anonymous):

\[du f.w=9/4\sqrt{5} fx(1,2) + 12/4\sqrt{5}(fy(1,2)\]

5 years ago
OpenStudy (anonymous):

because it need to use the magnitude for the vector also ....

5 years ago
OpenStudy (turingtest):

the magnitude of what? also, are you sure w is not 1/5sqrt(2)(3,4,5), because that would make it a unit vector, which I see now it is not...

5 years ago
OpenStudy (anonymous):

yes you are right as the exaple given as unit vector only.

5 years ago
OpenStudy (turingtest):

so then it should be\[\nabla f\cdot\vec w=\frac1{5\sqrt2}\langle12,14\sqrt3-12,0\rangle\cdot\langle3,4,5\rangle=\frac1{5\sqrt2}(36+56\sqrt3-48+0)\]

5 years ago