Question

Find the directions in which the directional derivative of \(f(x,y) - {x^2} + \sin xy\) at the point \((1,0)\) has the value 1

Short answer

The direction in which the directional derivative of the function \(f(x,y) = {x^2} + \sin xy\) at the point \((1,0)\) is 1 is \(\frac{{(2,1)}}{{\sqrt 5 }}\).

Step-by-step solution

Theorem used

Suppose\(f\)is a differentiable function of two or three variables. The maximum value of the directional derivative\({D_u}f(x)\)is\(|\nabla f(x)|\)and it occurs when u has the same direction as the gradient vector\(\nabla f(x){\rm{ }}\)

Calculate the rate of maximum change in directional derivative

It is given that the function is \(f(x,y) = {x^2} + \sin xy\) and the point is \((1,0)\).

The directional derivative of the function \(f(x,y) = {x^2} + \sin xy\) at the point \((1,0)\) is 1 .

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

Cbtain the value of \(\frac{{zy}}{{\partial x}}\) as follows.

\(\begin{aligned}{l}\frac{\partial }{{dx}}f(x,y) = \frac{\partial }{{dx}}\left( {{x^2} + \sin xy} \right)\\ = 2x + y\cos xy{\left. {\frac{\partial }{{dx}}f(x,y)} \right|_{x - 1,y = 0}}\\ = 2(1) + (0)\cos (1 - 0)\\ = 2\end{aligned}\)

Obtain the value of \(\frac{{ay}}{{2y}}\) as follows.

\(\begin{aligned}{l}\frac{\partial }{{\partial y}}f(x,y) = \frac{\partial }{{\partial y}}\left( {{x^2} + \sin xy} \right)\\ = x\cos xy{\left. {\frac{\partial }{{dy}}f(x,y)} \right|_{s = 1,y = 0}}\\ = (1)\cos (1 - 0) = 1\end{aligned}\)

Thus, the gradient of the function is \(\nabla f(x,y) = \langle 2,1\rangle \) -

The maximum value of the directional derivative is \(|\nabla f(x,y)|\) obtained as

\(\begin{aligned}{l}|\nabla f(x,y)| = |(2,1)|\\ = \sqrt {{{(2)}^2} + {{(1)}^2}} \;\;\\{\kern 1pt} = \sqrt {4 + 1} \\\;\;{\kern 1pt} = \sqrt 5 \end{aligned}\)

Thus, the related unit vector becomes \(\frac{{{\rm{v}}f(x,y)}}{{Nf(x,v)}} = \frac{{(\{ ,1)}}{{\sqrt 5 }}\) at which the \({D_0}f(x,y) = 1\).

Therefore, the direction in which the directional derivative of the function \(f(x,y) = {x^2} + \sin xy\) at the point \((1,0)\) is 1 is \(\frac{{\{ 2,1\} }}{{\sqrt 5 }}\).