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 $\vec u=2i+3j-5k$?

  • $\begingroup$ 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

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

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

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