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Chapter 12: Problem 49

Find an equation of the plane tangent to the following surfaces at the given point. $$\sin x y z=\frac{1}{2} ;\left(\pi, 1, \frac{1}{6}\right)$$

Short Answer

Expert verified
Answer: The equation of the tangent plane is \(-\frac{1}{6} (x - \pi) - \frac{\pi}{6} (y - 1) - \pi (z - \frac{1}{6}) = 0\).

Step by step solution

01

Find the gradient of the function

We need to compute the partial derivatives of the function with respect to x, y, and z: $$F(x,y,z) = \sin x y z$$ Now, find the partial derivatives of F: $$\frac{\partial F}{\partial x} = \cos x y z \cdot yz$$ $$\frac{\partial F}{\partial y} = \cos x y z \cdot xz$$ $$\frac{\partial F}{\partial z} = \cos x y z \cdot xy$$
02

Evaluate the partial derivatives at the given point

Now, evaluate the partial derivatives at the given point, \((\pi, 1, \frac{1}{6})\): $$\frac{\partial F}{\partial x}(\pi, 1, \frac{1}{6}) = \cos (\pi \cdot 1 \cdot \frac{1}{6}) \cdot (1 \cdot \frac{1}{6}) = -\frac{1}{6}$$ $$\frac{\partial F}{\partial y}(\pi, 1, \frac{1}{6}) = \cos (\pi \cdot 1 \cdot \frac{1}{6}) \cdot (\pi \cdot \frac{1}{6}) = -\frac{\pi}{6}$$ $$\frac{\partial F}{\partial z}(\pi, 1, \frac{1}{6}) = \cos (\pi \cdot 1 \cdot \frac{1}{6}) \cdot (\pi \cdot 1) = -\pi$$
03

Calculate the normal vector to the tangent plane

To calculate the normal vector to the tangent plane, we use the gradient components as the coordinates of the vector: $$\vec{n} = \langle -\frac{1}{6}, -\frac{\pi}{6}, -\pi \rangle$$
04

Formulate the equation of the tangent plane

Now, we can formulate the equation of the tangent plane using the point-normal form: $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ Substitute the components of the normal vector and the coordinates of the point: $$-\frac{1}{6} (x - \pi) - \frac{\pi}{6} (y - 1) - \pi (z - \frac{1}{6}) = 0$$ Finally, we have the equation of the tangent plane to the given surface at the specified point: $$-\frac{1}{6} (x - \pi) - \frac{\pi}{6} (y - 1) - \pi (z - \frac{1}{6}) = 0$$.

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