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Chapter 11: Problem 17

Find the gradient of the function at the given point. $$ z=\cos \left(x^{2}+y^{2}\right), \quad(3,-4) $$

Short Answer

Expert verified
The gradient of the function \(z=\cos \left(x^{2}+y^{2}\right)\) at the point (3,-4) is \((-6\sin(25),8\sin(25))\).

Step by step solution

01

Find the partial derivative with respect to x

The partial derivative of the function \(z=\cos \left(x^{2}+y^{2}\right)\) with respect to x is given by: \(\frac{\partial z}{\partial x} = -2x\sin(x^{2} + y^{2})\). Now plug in \(x=3\) and \(y=-4\) to get the partial derivative with respect to x at the given point.
02

Find the partial derivative with respect to y

Similarly, the partial derivative of the function \(z=\cos \left(x^{2}+y^{2}\right)\) with respect to y is given by: \(\frac{\partial z}{\partial y} = -2y\sin(x^{2} + y^{2})\). Now plug in \(x=3\) and \(y=-4\) to get the partial derivative with respect to y at the given point.
03

Put all together

Now we can put it all together. The gradient of the function at the point (3,-4) is given by the vector whose components are the partial derivatives with respect to x and y evaluated at the point (3,-4), i.e., \(\nabla z = (\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y})\).

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