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Chapter 11: Problem 16

Find the gradient of the function at the given point. $$ g(x, y)=2 x e^{y / x}, \quad(2,0) $$

Short Answer

Expert verified
The gradient of the function at the point (2, 0) is <2,4>.

Step by step solution

01

Calculate the partial derivatives

The first step is to compute the partial derivatives of \(g\) with respect to \(x\) and \(y\). \(\frac{\partial g}{\partial x} = e^{y/x}(2 - y/x)\) \(\frac{\partial g}{\partial y} = 2xe^{y/x}\)
02

Evaluate the partial derivatives at the point

The second step is to evaluate the partial derivatives at the point (2, 0). Using the point (2,0) in \(\frac{\partial g}{\partial x}\): \(\frac{\partial g}{\partial x} = e^{0/2}(2 - 0/2) = 2\)Using the point (2,0) in \(\frac{\partial g}{\partial y}\): \(\frac{\partial g}{\partial y} = 2*2*e^{0/2} = 4\)
03

Express the Gradient

Finally, express the gradient. The gradient of the function at the point is given by the vector of these two values as done in the following step: \(\nabla g(2,0) = <2,4>\)

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Most popular questions from this chapter

Show that \(\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}=0\) for \(w=f(x, y), x=u-v,\) and \(y=v-u\)

Find \(\partial w / \partial s\) and \(\partial w / \partial t\) by using the appropriate Chain Rule. \(w=z e^{x / y}, \quad x=s-t, \quad y=s+t, \quad z=s t\)

Define the derivative of the function \(z=f(x, y)\) in the direction \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\).

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y z, \quad x=t^{2}, \quad y=2 t, \quad z=e^{-t}\)

If \(f(x, y)=0,\) give the rule for finding \(d y / d x\) implicitly. If \(f(x, y, z)=0,\) give the rule for finding \(\partial z / \partial x\) and \(\partial z / \partial y\) implicitly.
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