Problem 13 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$x y \sin z=1 ;(1,2, \pi / 6) \text { and }(-2,-1,5 \pi / 6)$$

Step-by-Step Solution

Answer

Answer: The equations of the tangent planes at the given points are: At point (1,2,π/6): $$x - \frac{1}{2}y + \sqrt{3}z = 1 + \sqrt{3}\frac{\pi}{6}$$ At point (-2,-1, 5π/6): $$-\frac{1}{2}x - y + \sqrt{3}z = -2 - \frac{1}{2} - \sqrt{3}\frac{5\pi}{6}$$

Step 1: Find the gradient of the surface

First, we need to find the partial derivatives of the given function: $$F(x, y, z) = xy\sin(z) - 1$$ With respect to x, y, and z: $$F_x = \frac{\partial F}{\partial x} = y\sin(z)$$ $$F_y = \frac{\partial F}{\partial y} = x\sin(z)$$ $$F_z = \frac{\partial F}{\partial z} = xy\cos(z)$$ Now we can evaluate the gradient at each given point: At point (1,2,π/6): $$\nabla F(1, 2, \frac{\pi}{6}) = (2\sin(\frac{\pi}{6}), 1\sin(\frac{\pi}{6}), 1\cdot2\cos(\frac{\pi}{6})) = (1, \frac{1}{2}, \sqrt{3})$$ At point (-2,-1, 5π/6): $$\nabla F(-2, -1, \frac{5\pi}{6}) = (-1\sin(\frac{5\pi}{6}), -2\sin(\frac{5\pi}{6}), -2\cdot(-1)\cos(\frac{5\pi}{6})) = (-\frac{1}{2}, -1, \sqrt{3})$$

Step 2: Use the point-normal form for a plane

Now we can use the point-normal form of a plane to find the equation of the tangent planes at each point. The point-normal form is given by: $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ Where (x_0, y_0, z_0) is the tangent point, and (A, B, C) is the normal vector (gradient). At point (1,2,π/6): $$1 (x-1) + \frac{1}{2}(y-2) + \sqrt{3}(z - \frac{\pi}{6}) = 0$$ At point (-2,-1, 5π/6): $$-\frac{1}{2} (x + 2) - 1(y + 1) + \sqrt{3}(z - \frac{5\pi}{6}) = 0$$ To conclude, the equation of the tangent plane at point (1,2,π/6) is: $$x - \frac{1}{2}y + \sqrt{3}z = 1 + \sqrt{3}\frac{\pi}{6}$$ And the equation of the tangent plane at point (-2,-1, 5π/6) is: $$-\frac{1}{2}x - y + \sqrt{3}z = -2 - \frac{1}{2} - \sqrt{3}\frac{5\pi}{6}$$

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how functions of multiple variables change. They allow you to analyze how each variable, independently of others, contributes to the rate of change of the function. In this problem, the function is given by \[ F(x, y, z) = xy\sin(z) - 1 \] To compute the partial derivatives, consider each variable one at a time while treating others as constants. This results in:

• \( F_x = \frac{\partial F}{\partial x} = y\sin(z) \)
• \( F_y = \frac{\partial F}{\partial y} = x\sin(z) \)
• \( F_z = \frac{\partial F}{\partial z} = xy\cos(z) \)

These derivatives form the components of the gradient vector, which indicates the direction of the steepest increase of the function. The gradient plays a critical role, especially in determining the normal vector to a surface, which is essential in finding tangent planes.

Point-Normal Form

The point-normal form of a plane equation is a neat way to express the relationship between any point on the plane and the normal vector. A plane can be defined by a single point on the plane, \( (x_0, y_0, z_0) \), and a vector perpendicular to the plane, known as the normal vector \( (A, B, C) \).The point-normal form equation is:\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]This equation highlights that any point \( (x, y, z) \) on the plane will satisfy the relation as long as the dot product of the vector formed by \( (x, y, z) - (x_0, y_0, z_0) \) with the normal vector \( (A, B, C) \) is zero.Through this exercise, you apply this concept to find the equations of tangent planes at the given points \( (1, 2, \pi/6) \) and \( (-2, -1, 5\pi/6) \) using the previously calculated gradient as the normal vector.

Equations of Surfaces

Surface equations represent the relationship between points that lie on a surface in three-dimensional space. These equations are typically given in the form \( f(x, y, z) = 0 \). For any point \((x, y, z)\) that lies on the surface, swapping the variables into the function will satisfy the equation.In this context, the surface is defined as:\[ F(x, y, z) = xy\sin(z) - 1 = 0 \]This equation describes a complex surface depending both on trigonometric functions and the interaction between variables. For each given point to find the tangent plane, it is essential first to compute the gradient, which acts as a normal to the surface, and then apply the point-normal form.Understanding these equations is crucial, as they not only describe surfaces but also aid in visualizing and manipulating geometric shapes and planes in higher dimensions.