Question

Find the directional derivative of the function at the
given point in the direction of the vector\(v\).

\(\left. {g(r,s) = {{{\mathop{\rm lin}\nolimits} }^{ - 1}}(r3),(1,2),\quad r = 51 + 10} \right\}\)

Short answer

The directional derivative is \(\nabla g(1,2) = \frac{4}{{5\sqrt 5 }}\).

Step-by-step solution

using the directional derivative

It is given that the function is \(g(r,s) = {\tan ^{ - 1}}(rs)\), the point is \((1,2)\) and the directed vector is \({\bf{v}} = 5{\bf{i}} + 10{\bf{j}} = \langle 5,10\rangle \)

The directional derivative of the function\(f\)at the given point in the direction of the unit vector\({\bf{u}}\)is given by\({D_{\bf{u}}}f(x,y) = \left\langle {{f_x},{f_y}} \right\rangle \cdot {\bf{u}}\).

The unit vector becomes

\(\begin{aligned}{l}{\bf{u}} = \frac{{5{\bf{l}} + 10{\bf{j}}}}{{\sqrt {{5^2} + {{10}^2}} }}\\ = \frac{{5{\bf{l}} + 10{\bf{j}}}}{{\sqrt {125} }}\\ = \frac{{\langle 5,10\rangle }}{{5\sqrt 5 }}\end{aligned}\)

Find the value with respect to r

Obtain the value of \(\frac{{\partial g}}{{\partial r}}\) as follows.

\(\begin{aligned}{l}\frac{\partial }{{\partial r}}g(r,s) = \frac{\partial }{{\partial r}}\left( {{{\tan }^{ - 1}}(rs)} \right)\\ = \frac{1}{{1 + {{(rs)}^2}}}\\\frac{\partial }{{\partial r}}(rs) = \frac{s}{{1 + {{(rs)}^2}}}\\\frac{\partial }{{\partial p}}g(1,2) = \frac{{(2)}}{{1 + {{(1 - 2)}^2}}}\\ = \frac{2}{{1 + 4}}\\ = \frac{2}{5}\end{aligned}\)

Find the value with respect to s

Obtain the value of \(\frac{{\partial g}}{{\partial s}}\) as follows.

\(\begin{aligned}{l}\frac{\partial }{{\partial s}}g(r,s) = \frac{\partial }{{\partial s}}\left( {{{\tan }^{ - 1}}(rs)} \right)\\ = \frac{1}{{1 + {{(rs)}^2}}}\frac{\partial }{{\partial s}}(rs)\\ = \frac{r}{{1 + {{(rs)}^2}}}\\\frac{\partial }{{\partial p}}g(1,2) = \frac{{(1)}}{{1 + {{(1 - 2)}^2}}}\\ = \frac{1}{{1 + 4}}\\ = \frac{1}{5}\end{aligned}\)

use the directional derivative with unit vector

Thus, the directional derivative along the unit vector \({\bf{u}} = \frac{{\langle 5,10\rangle }}{{5\sqrt 5 }}\) becomes

\(\begin{aligned}{l}{D_u}f(2,1) = \left\langle {\frac{2}{5},\frac{1}{5}} \right\rangle \cdot \frac{{(5,10\rangle }}{{5\sqrt 5 }}\\ = \frac{1}{{5\sqrt 5 }}\left( {\left( {\frac{2}{5}} \right)(5) + \left( {\frac{1}{5}} \right)(10)} \right)\\ = \frac{1}{{5\sqrt 5 }}(2 + 2)\\ = \frac{4}{{5\sqrt 5 }}\end{aligned}\)

Hence, the directional derivative of the function at the given point in the direction along the given vector is \({D_{\rm{u}}}f(1,2) = \frac{4}{{5\sqrt 5 }}\).