Question

Find the maximum rate of change of\(f\)at the given point and the direction in which it occurs.

\(f(s,t) = t{e^{nt}},\quad (0,2)\)

Short answer

The maximum rate of change of \(f(s,t) = t{e^{st}}\) at the point \((0,2)\) is \(|\nabla f(0,2)| = \sqrt {17} \) and it occurs in the direction \(\langle 4,1\rangle \).

Step-by-step solution

Formula 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\(\nabla f{(x)^{\prime \prime }}\)

The maximum rate of change of \(f(s,t) = t{e^{st}}\) at the point \((0,2)\) is computed as follows,

\(\begin{aligned}{l}\nabla f(s,t) = \left\langle {{f_s},{f_t}} \right\rangle \\ = \left\langle {\frac{\partial }{{\partial s}}\left( {t{e^{st}}} \right),\frac{\partial }{{\partial t}}\left( {t{e^{st}}} \right)} \right\rangle \\ = \left\langle {\left( {{t^2}{e^{st}}} \right),\left( {{e^{st}} + st{e^{st}}} \right)} \right\rangle \\ = \left\langle {\left( {{t^2}{e^{st}}} \right),(1 + st){e^{st}}} \right\rangle \end{aligned}\)

Thus, \(\nabla f(s,t) = \left\langle {\left( {{t^2}{e^{st}}} \right),(1 + st){e^{st}}} \right\rangle \).

Maximum rate of change

Substitute \(s = 0,t = 2\) in \(\nabla f(s,t)\) and find the value of \(\nabla f(0,2)\).

Thus, the maximum rate of change occurs in the direction \(\langle 4,1\rangle \).

The maximum rate is calculated as follows,

\(\begin{aligned}{l}|\nabla f(0,2)| = \sqrt {{4^2} + {1^2}} \\ = \sqrt {16 + 1} \\ = \sqrt {17} \end{aligned}\)

Thus, the maximum rate is \(|\nabla f(4,1)| = \sqrt {17} \) which occurs in the direction \(\langle 4,1\rangle \).