Question

Let f(P)=16 and f(Q)=24 where P=(4,5) and Q=(4.03,4.96)
Approximate the directional derivative of f at P in the direction of Q.

Answer 1

If it helps, I've found the magnitudes of P and Q.
\[|P|=(4/\sqrt{41},5/\sqrt{41}), |Q|=(0.63,0.78)\]

Answer 2

it looks to me like they want you to use linear approximation or differentials, since \(P\approx Q\)

Answer 3

USE limits of multivariable functions to solve this, i.e use partial derivative definition of limit. Directional is given by "nabla" the inverted trianlge operator

Answer 4

\[\lim_{h \rightarrow 0}\frac{ f(a+h,b)-f(a,b) }{ h }\]\[\lim_{h \rightarrow 0}\frac{ f(4+h,5)-f(4,5) }{ h }\]\[\lim_{h \rightarrow 0}\frac{ f(4+h,5)-16 }{ h }\]

Answer 5

for vectors \(Q\approx P\), we have the multivariable linear approximation as\[D_{\vec u}f(x,y)\approx f(P)+\nabla f(P)\cdot(Q-P)\]

Answer 6

I'm not sure how to find the gradient of f(P), since it's a whole number.
\[D_u \approx16+gradf(P)*(0.03,-0.04)\]

Answer 7

Oh I messed up...
I am sort of winging it here but I was supposed to write\[f(Q)\approx f(P)+\nabla f(P)\cdot(Q-P)\]So I wonder if we can get to something like\[f(Q)\approx f(P)+D_{\vec Q}f(P)(Q-P)\]somehow, the, solve for \(D_{\vec Q}f(P)\)
not sure about the second formula though; I sort of just made it up...

Answer 8

\[24=16+∇f(P)⋅(0.03,-0.04)\]\[8=∇f(P)⋅(0.03,−0.04)\]Not even sure if I can do that, but I did.

Is your dot in there symbolizing a dot product, or multiplication? I'm assuming it's a dot product.