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Chapter 11: Problem 12

Find the directional derivative of the function in the direction of \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\). $$ g(x, y)=x e^{y}, \quad \theta=\frac{2 \pi}{3} $$

Short Answer

Expert verified
Therefore, the directional derivative of the function \(g(x, y) = x e^{y}\) in the direction of the vector \(\mathbf{u}\) is \(\nabla_{\mathbf{u}} g = -\frac{e^{y}}{2} + \frac{\sqrt{3} x e^{y}}{2}\).

Step by step solution

01

Calculate the Gradient of the Function

The first step is finding the gradient of the given function. This requires differentiating \(g\) with respect to \(x\) and \(y\). The gradient of function \(g(x, y) = x e^{y}\) is given by \(\nabla g =\left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right)\). On differentiating, we get \(\nabla g = \left(e^{y}, x e^{y}\right)\).
02

Calculate the Direction Vector

To find the direction vector using \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\) where \(\theta=\frac{2 \pi}{3}\), we replace \(\theta\) with the given value and get \(\mathbf{u} = \left( \cos \left(\frac{2 \pi}{3}\right), \sin \left(\frac{2 \pi}{3}\right)\right) = \left(-\frac{1}{2},\sqrt{3}/2\right)\).
03

Calculate the Directional Derivative

Finally, to compute the directional derivative of \(g\) at a point \((x,y)\), in the direction of \(\mathbf{u}\), we use the formula \(\nabla_{\mathbf{u}} g = \nabla g\cdot\mathbf{u}\). Substituting the appropriate values, we get \[\nabla_{\mathbf{u}} g = \nabla g\cdot\mathbf{u} = \left(e^{y}, x e^{y}\right) \cdot \left(-\frac{1}{2},\sqrt{3}/2\right) = -\frac{e^{y}}{2} + \frac{\sqrt{3} x e^{y}}{2}\].

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Most popular questions from this chapter

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