0
$\begingroup$

Do we only calculate directional derivatives for scalar-valued functions?

Is it not possible to calculate directional derivatives for vector-valued functions?

How about using the vector of directional derivatives of the components of the given vector function? Would there be any useful physical or geometric meaning?

For a specific (randomly chosen) example, if $\vec v(x,y,z)$ is given by $$ \vec v(x,y,z)= \begin{bmatrix} x^3+y^2+z\\ ze^x\\ xyz-9xz\\ \end{bmatrix} $$ how can we interpret the directional derivative of $\vec v$ at the point $(1,2,3)$ in the direction of the vector $\vec u=2i+3j-5k$?

$\endgroup$
  • $\begingroup$ Just compute the directional derivative of each of the three components separately. The geometric meaning is that this vector tells you the infinitesimal change of your vector field. $\endgroup$ – Nick Nov 14 '18 at 17:03
  • $\begingroup$ would you please elaborate your answer? I want to see some computation.. $\endgroup$ – Rakibul Islam Prince Nov 14 '18 at 19:52
1
$\begingroup$

The generic formula for the directional derivative of a function $f$ in the direction $u$ (for a unit vector) is $D_u f (x,y,z) = \nabla f(x,y,z) \cdot u$. For a vector, just do this to all the components.

Let's look at the example you give. Let's call $f_1(x,y,z) = x^3+y^2+z$, $\, f_2(x,y,z) = ze^x$, and $f_3(x,y,z) = xyz - 9xz$. Then the gradients are

$$ \begin {align*} \nabla f_1 &= (3x^2, \, 2y, \, 1) \\ \nabla f_2 &= (ze^x, \, 0, \, e^x) \\ \nabla f_3 &= (yz - 9z, \, xz, \, xy - 9x) \end {align*} $$

At your particular point $(1,2,3)$, these are:

$$ \begin {align*} \nabla f_1(1,2,3) &= (3,4,1) \\ \nabla f_2(1,2,3) &= (3e,0,e) \\ \nabla f_3(1,2,3) &= (-27,3,-7) \end {align*} $$

The formula I mentioned above for directional derivative requires a UNIT vector. Since the vector you give, $u = (2,3,-5)$ is NOT a unit vector, we have to rescale it, and instead use the vector

$$ w = \frac{1}{\|u\|}u = \frac{1}{\sqrt{38}}(2,3,-5) $$

Now finally use the formula:

$$ \begin {align*} D_w f_1(1,2,3) &= \frac{1}{\sqrt{38}}(2,3,-5) \cdot (3,4,1) = \frac{13}{\sqrt{38}} \\ D_w f_2(1,2,3) &= \frac{1}{\sqrt{38}}(2,3,-5) \cdot (3e,0,e) = \frac{e}{\sqrt{38}} \\ D_w f_3(1,2,3) &= \frac{1}{\sqrt{38}}(2,3,-5) \cdot (-27,3,-7) = \frac{98}{\sqrt{38}} \end {align*} $$

Putting it all together, the directional derivative of your vector function $v(x,y,z)$ in the direction u (also the same as direction w) is given by

$$ \frac{1}{\sqrt{38}} \left( \begin {array}{c} 13 \\ e \\ 98 \end{array} \right) $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.