Problem 17 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$z=4-2 x^{2}-y^{2} ;(2,2,-8) \text { and }(-1,-1,1)$$

Step-by-Step Solution

Answer

Question: Find the equations of the tangent planes for the surface $$z = 4 - 2x^2 - y^2$$ at the points (2,2,-8) and (-1,-1,1). Answer: The equations of the tangent planes at points (2,2,-8) and (-1,-1,1) are: $$-8x - 4y + 24 = 0$$ $$4x + 2y + 6 = 0$$

Step 1: Determine the partial derivatives

First, we need to find the partial derivatives of the function $$z = 4 - 2x^2 - y^2$$. Take the derivative with respect to x and y, respectively: $$\frac{\partial z}{\partial x} = -4x $$ $$\frac{\partial z}{\partial y} = -2y $$

Step 2: Calculate the gradients at each point

Next, calculate the gradient (normal vectors) of the surface at each point by evaluating the partial derivatives at each of the points: For point (2,2,-8): $$\nabla z_1 = \left(-4(2), -2(2)\right) = (-8,-4)$$ For point (-1,-1,1): $$\nabla z_2 = \left(-4(-1), -2(-1)\right) = (4,2)$$

Step 3: Use the normal vectors to write the equation of the tangent planes

Now, with the normal vectors (gradients) calculated, we can write the equation of the tangent planes at each point using the point-normal form of the plane equation, which is given by: $$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$$ where (a,b,c) are the components of the normal vector, and (x0,y0,z0) are the coordinates of the given point. For point (2,2,-8) and normal vector (-8,-4,0): $$-8(x-2) - 4(y-2) + 0(z+8) = 0$$ Simplify the equation: $$-8x +16 - 4y +8 = 0$$ $$-8x - 4y + 24 = 0$$ For point (-1,-1,1) and normal vector (4,2,0): $$4(x+1) + 2(y+1) + 0(z-1) = 0$$ Simplify the equation: $$4x +4 +2y +2 = 0$$ $$4x + 2y + 6 = 0$$

Step 4: Final tangent plane equations

The final equations of the tangent planes at points (2,2,-8) and (-1,-1,1) are: $$-8x - 4y + 24 = 0$$ $$4x + 2y + 6 = 0$$

Key Concepts

Partial Derivatives

Partial derivatives are crucial in finding the tangent plane to a surface. They involve taking the derivative of a multivariable function with respect to one variable, holding the others constant. In this context, for the surface defined by the function \( z = 4 - 2x^2 - y^2 \), we need to find how \( z \) changes as \( x \) and \( y \) vary.

The partial derivative with respect to \( x \), \( \frac{\partial z}{\partial x} \), captures the rate of change of \( z \) with respect to \( x \) alone. Similarly, \( \frac{\partial z}{\partial y} \) shows the change in \( z \) as \( y \) varies. The calculated partial derivatives are:

• \( \frac{\partial z}{\partial x} = -4x \)
• \( \frac{\partial z}{\partial y} = -2y \)

These derivatives help determine the slope of the tangent plane at any given point on the surface.

Gradient Calculations

Having the partial derivatives, we can now move on to gradient calculations. The gradient of a scalar field, denoted by \( abla z \), is a vector pointing in the direction of the greatest increase of the function. For a surface \( z = f(x,y) \), the gradient is given by:

\( abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) \)

At different points on the surface, the gradient might vary. For this exercise, calculating \( abla z \) at the points (2,2,-8) and (-1,-1,1) provides normal vectors for these points:

• At point (2,2,-8), \( abla z_1 = (-8, -4) \)
• At point (-1,-1,1), \( abla z_2 = (4, 2) \)

These vectors are essential in formulating the tangent plane equations.

Point-Normal Form

The point-normal form of a plane equation expresses how a plane is oriented in space by using a point on the plane and a normal vector perpendicular to it. The general equation for a plane given a normal vector \( (a, b, c) \) and a point \( (x_0, y_0, z_0) \) is:

\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]

For our specific exercise, each normal vector (obtained from the gradient) and the corresponding point allows us to establish the equation of the tangent plane at that point. These normal vectors \( (-8, -4, 0) \) and \( (4, 2, 0) \) show that the planes are horizontal with respect to the \( z \)-axis, confirming that the surface slope is consistent with changes in \( x \) and \( y \).

Plane Equation

With both the normal vector and a specific point, we can derive the equation of the tangent plane, which provides valuable geometric interpretations of the surface at that point. For this exercise, we have:

For point (2,2,-8) with normal vector \((-8,-4,0)\):

• \(-8(x-2) - 4(y-2) = 0\) simplifies to \(-8x - 4y + 24 = 0\)

For point (-1,-1,1) with normal vector \((4,2,0)\):

• \(4(x+1) + 2(y+1) = 0\) simplifies to \(4x + 2y + 6 = 0\)

These equations illustrate the planes tangent to the surface at specified points, using the calculated gradients as normal vectors. They represent the spatial position and orientation of the tangent planes.