Question

(a) Find the gradient of \(f\).

(b) Evaluate the gradient at the point \(P\).

(c) Find the rate of change of \(f\) at \(P\) in the direction of the vector \({\bf{u}}\).

\(f(x,y) = \sin (2x + 3y),\quad P( - 6,4),\quad {\bf{u}} = \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\)

Short answer

(a) The gradient of the given function\(f(x,y) = \sin (2x + 3y)\)is\(\langle 2\cos (2x + 3y),3\cos (2x + 3y)\rangle \).

(b) The gradient\(\nabla f\)at the point\(P( - 6,4)\)is\(\langle 2,3\rangle \).

(c) The directional derivative of the function \(f\) at the given point in the direction of the vector \({\bf{u}} = \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\) is \(\underline {\sqrt 3 - \frac{3}{2}} .\)

Step-by-step solution

 The gradient and directional derivative of a function

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

The gradient of the function

It is given that the function is\(f(x,y) = \sin (2x + 3y)\), the point is\(P( - 6,4)\)and the unit vector is\({\bf{u}} = \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\).

Obtain the value of\(\nabla f\)as follows.

\(\begin{aligned}{l}\nabla f(x,y) = \left\langle {\frac{\partial }{{\partial x}}\sin (2x + 3y),\frac{\partial }{{\partial y}}\sin (2x + 3y)} \right\rangle \\ = \langle \cos (2x + 3y)(2),\cos (2x + 3y)(3)\rangle \\ = \langle 2\cos (2x + 3y),3\cos (2x + 3y)\rangle \end{aligned}\)

Hence, the gradient of the given function \(f(x,y) = \sin (2x + 3y)\) is \(\langle 2\cos (2x + 3y),3\cos (2x + 3y)\rangle \).

The gradient of function at given point

(b)

From part a, the gradient of the given function\(f(x,y) = \sin (2x + 3y)\)is\(\langle 2\cos (2x + 3y),3\cos (2x + 3y)\rangle \).

At point\(P( - 6,4)\), the gradient is obtained as

\(\begin{aligned}{l}\nabla f( - 6,4) = \langle 2\cos (2( - 6) + 3(4)),3\cos (2( - 6) + 3(4))\rangle \\ = \langle 2\cos 0,3\cos 0\rangle \\ = \langle 2(1),3(1)\rangle \\ = \langle 2,3\rangle \end{aligned}\)

Hence, the gradient\(\nabla f\)at the point\(P( - 6,4)\)is\(\langle 2,3\rangle \).

The directional derivative of the function at given point

(c)

From part b, the gradient\(\nabla f\)at the point\(P( - 6,4)\)is\(\langle 2,3\rangle \)that is\(\frac{{\partial f}}{{\partial x}}( - 6,4) = 2\)and\(\frac{{\partial f}}{{\partial x}}( - 6,4) = 3\).

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

Thus, the directional derivative along the unit vector is\({\bf{u}} = \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\)becomes

\(\begin{aligned}{l}{D_{\bf{u}}}f(2,3) = \langle 2,3\rangle \cdot \left\langle {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right\rangle \\ = 2 \cdot \frac{{\sqrt 3 }}{2} + 3 \cdot \left( { - \frac{1}{2}} \right)\\ = \sqrt 3 - \frac{3}{2}\end{aligned}\)

Hence, the directional derivative of the function \(f\) at the given point in the direction of the vector \({\bf{u}} = \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\) is \(\underline {\sqrt 3 - \frac{3}{2}} \).