Question

Find the directional derivative of \(f\) at the given point in the direction indicated by the angle \(\theta \).

\(f(x,y) = {x^3}{y^4} + {x^4}{y^3},\quad (1,1),\quad \theta = \pi /6\)

Short answer

The directional derivative of the function \(f\) at a point \((1,1)\) in the direction indicated by the angle \(\frac{\pi }{6}\) is \({D_{\bf{u}}}f(1,1) = \frac{7}{2}(\sqrt 3 + 1)\).

Step-by-step solution

The directional derivative of a function

The directional derivative of the function\(f\)at the given point in the direction indicated by the angle\(\theta \)such that the unit vector is\({\bf{u}} = \langle \cos \theta ,\sin \theta \rangle \)is given by

\({D_{\bf{u}}}f(x,y) = {f_x}(x,y)\cos \theta + {f_y}(x,y)\sin \theta \).

Obtain the value of \(\frac{{\partial f}}{{\partial x}}\) 

Obtain the value of\(\frac{{\partial f}}{{\partial x}}\)as follows.

\(\begin{aligned}{l}\frac{\partial }{{\partial x}}f(x,y) = \frac{\partial }{{\partial x}}\left( {{x^3}{y^4} + {x^4}{y^3}} \right)\\ = {y^4}\frac{\partial }{{\partial x}}{x^3} + {y^3}\frac{\partial }{{\partial x}}{x^4}\\ = {y^4}\left( {3{x^2}} \right) + {y^3}\left( {4{x^3}} \right)\\ = 3{x^2}{y^4} + 4{x^3}{y^3}\\\frac{\partial }{{\partial x}}f(1,1) = 3{(1)^2}{(1)^4} + 4{(1)^3}{(1)^3}\\ = 7\end{aligned}\)

Obtain the value of\(\frac{{\partial f}}{{\partial y}}\)as follows.

\(\begin{aligned}{l}\frac{\partial }{{\partial y}}f(x,y) = \frac{\partial }{{\partial y}}\left( {{x^3}{y^4} + {x^4}{y^3}} \right)\\ = {x^3}\frac{\partial }{{\partial y}}{y^4} + {x^4}\frac{\partial }{{\partial y}}{y^3}\\ = {x^3}\left( {4{y^3}} \right) + {x^4}\left( {3{y^2}} \right)\\ = 4{x^3}{y^3} + 3{x^4}{y^2}\\\frac{\partial }{{\partial y}}f(1,1) = 4{(1)^3}{(1)^3} + 3{(1)^4}{(1)^2}\\ = 7\end{aligned}\)

The directional derivative of the function

Thus, the directional derivative at an angle\(\theta = \frac{\pi }{6}\)becomes

\(\begin{aligned}{l}{D_{\bf{u}}}f(1,1) = {f_x}(1,1)\cos \frac{\pi }{6} + {f_y}(1,1)\sin \frac{\pi }{6}\\ = (7)\left( {\frac{{\sqrt 3 }}{2}} \right) + (7)\left( {\frac{1}{2}} \right)\\ = \frac{7}{2}(\sqrt 3 + 1)\end{aligned}\)

That is, \({D_{\bf{u}}}f(1,1) = \frac{7}{2}(\sqrt 3 + 1)\).