Problem 49 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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 1: Find the Gradient of the Given Surface

To find the normal vector to the surface, first, we need to find the gradient of the given function. Let's call the function F(x, y, z): $$F(x, y, z) = \sin x y z - \frac{1}{2} = 0$$. Now, we find the gradient as follows: $$\nabla F(x, y, z) = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ Start by finding the partial derivative with respect to x: $$\frac{\partial F}{\partial x} = yz\cos x y z$$ Now, find the partial derivative with respect to y: $$\frac{\partial F}{\partial y} = xz\cos x y z$$ Finally, find the partial derivative with respect to z: $$\frac{\partial F}{\partial z} = xy\cos x y z$$ Thus, the gradient vector is: $$\nabla F(x, y, z) = \left(yz\cos x y z, xz\cos x y z, xy\cos x y z \right)$$

Step 2: Evaluate the Gradient at the Given Point

Now, we need to evaluate the gradient at the given point: $$\nabla F(\pi, 1, \frac{1}{6}) = \left(\frac{1}{6}\cos{\pi}, \frac{\pi}{6}\cos{\pi},\pi\cos{\pi}\right)$$ The cosine of \(\pi\) is -1, so the gradient is: $$\nabla F(\pi, 1, \frac{1}{6}) = \left(-\frac{1}{6}, -\frac{\pi}{6},-\pi\right)$$ This gradient vector is the normal vector to the tangent plane.

Step 3: Create the Equation of the Tangent Plane

Now, we can use the point-slope form of the equation of a plane with the given point and normal vector: $$A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$$ Using the normal vector components as coefficients and the given point \((\pi, 1, \frac{1}{6})\) as the point on the plane, we get the equation of the tangent plane: $$-\frac{1}{6}(x-\pi) -\frac{\pi}{6}(y-1) -\pi(z-\frac{1}{6}) = 0$$