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OpenStudy (anonymous):
Use the gradient to find the directional derivative of the function at P in the direction of Q.
f(x,y) = sin2xcosy, P(pi,0),Q(pi/2,pi)
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OpenStudy (anonymous):
For vector PQ I got -pi/2i-pi
OpenStudy (experimentx):
how did you get that?
OpenStudy (anonymous):
I took pi/2-pi , and pi-0 to get -pi/2i-pij
OpenStudy (experimentx):
yeah ...okay that right!!! i understand!!
OpenStudy (anonymous):
then would the unit vector where u=v/||v|| be -pi/2i-pij/[squarert(pi^2/4+pi^2)]?
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OpenStudy (experimentx):
yeah ... that would be correct!!
OpenStudy (experimentx):
okay, i think i get it
d(PQ)/dr = gradient of f(x,y) .dot. unit vector
OpenStudy (anonymous):
yeah thats the formula.
OpenStudy (experimentx):
you have the unit vector above ... the gradient is
i 2 cos(2x).cos(x) - j sin(2x).sin(x)
OpenStudy (anonymous):
yeah well i got thetaf(x,y)=2cos2xcosyi-sin2xsinyj
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OpenStudy (experimentx):
i haven't calculated, must be something like
-pi*cos(2x)(cosy)/|v|+pi*sin(2x) (sinx)/|v|
since it is a dot product, it should be scalar
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