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Chapter 11: Problem 20

Use the gradient to find the directional derivative of the function at \(P\) in the direction of \(Q\). $$ f(x, y)=\sin 2 x \cos y, \quad P(0,0), Q\left(\frac{\pi}{2}, \pi\right) $$

Short Answer

Expert verified
The directional derivative of the function \(f(x, y)=\sin 2x \cos y\) at \(P(0,0)\) in the direction toward \(Q\left(\frac{\pi}{2}, \pi\right)\) is \(D_{\mathbf{u}}f = \frac{2}{\sqrt{1^2 + (\pi)^2}}.\)

Step by step solution

01

Compute the Gradient of the function at \(P(0,0)\)

The gradient of function \(f\) is defined as \(\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). Here, \(f(x, y) = \sin 2x \cos y\). So, we need to find \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), which will return the vector \( \nabla f = \left(2 \cos 2x \cos y, -\sin 2x \sin y\right) \). The gradient at \(P(0,0)\) is thus \( \nabla f(0,0) = \left(2,0\right)\).
02

Compute the unit vector

A unit vector from \(P\) to \(Q\) is given by the formula \(\frac{Q-P}{|Q-P|}\), which would give us \(\frac{(\frac{\pi}{2}-0, \pi-0)}{\left|\left(\frac{\pi}{2}-0, \pi-0\right)\right|} = \left(\frac{1}{\sqrt{1^2 + (\pi)^2}}, \frac{\pi}{\sqrt{1^2 + (\pi)^2}}\right)\).
03

Find the dot product of the gradient and the unit vector

The directional derivative \(D_{\mathbf{u}}f\) in the direction of a unit vector \(\mathbf{u}\) is given by \(\nabla f \cdot \mathbf{u}\). Using the gradient at \(P(0,0)\) from Step 1 and the unit vector from Step 2, we find \(D_{\mathbf{u}}f = \left(2,0\right) \cdot \left(\frac{1}{\sqrt{1^2 + (\pi)^2}}, \frac{\pi}{\sqrt{1^2 + (\pi)^2}}\right) = \frac{2}{\sqrt{1^2 + (\pi)^2}}\).

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