Problem 12 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$x^{2}+y^{2}-z^{2}=0 ;(3,4,5) \text { and }(-4,-3,5)$$

Step-by-Step Solution

Answer

Answer: The equation of the tangent plane at point \((3, 4, 5)\) is \(6x + 8y - 10z = 38\), and the equation of the tangent plane at point \((-4, -3, 5)\) is \(-8x - 6y - 10z = -86\).

Step 1: Compute the gradient of the function

To find the gradient of the function \(f(x,y,z) = x^{2}+y^{2}-z^{2}\), we compute the partial derivatives with respect to \(x\), \(y\), and \(z\): $$\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = -2z$$ The gradient of the function is given by \(\nabla f = \left(2x, 2y, -2z\right)\).

Step 2: Evaluate the gradient at the given points

Now, we evaluate the gradient at the given points \((3, 4, 5)\) and \((-4, -3, 5)\). For \((3, 4, 5)\), we have: $$\nabla f(3, 4, 5) = \left(2 \cdot 3, 2 \cdot 4, -2 \cdot 5\right) = (6, 8, -10)$$ For \((-4, -3, 5)\), we have: $$\nabla f(-4, -3, 5) = \left(2 \cdot -4, 2 \cdot -3, -2 \cdot 5\right) = (-8, -6, -10)$$

Step 3: Find the equations of the tangent planes

Now, use the dot product formula and the given points to find the equations of the tangent planes. Recall that the dot product of the gradient and the normal vector is equal to the constant term in the equation of the tangent plane. For \((3, 4, 5)\): $$6(x - 3) + 8(y - 4) - 10(z - 5) = 0$$ Simplifying, we obtain: $$6x + 8y - 10z = 38$$ This is the equation of the tangent plane at point \((3, 4, 5)\). For \((-4, -3, 5)\): $$-8(x + 4) - 6(y + 3) - 10(z - 5) = 0$$ Simplifying, we obtain: $$-8x - 6y - 10z = -86$$ This is the equation of the tangent plane at point \((-4, -3, 5)\).

Key Concepts

Gradient

Understanding the gradient is essential in finding the equation of a tangent plane. The gradient, denoted by \( abla f \), is a vector that contains all the partial derivatives of a function. It provides essential information about the direction and rate of steepest ascent of the function. When dealing with surfaces, the gradient vector is perpendicular to the surface at a given point, turning it into a powerful tool for tangent plane calculations.

To compute the gradient of a function such as \( f(x, y, z) = x^{2} + y^{2} - z^{2} \), consider the partial derivatives with respect to each variable:

• Partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 2x \)
• Partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = 2y \)
• Partial derivative with respect to \( z \): \( \frac{\partial f}{\partial z} = -2z \)

The gradient vector is then \( abla f = (2x, 2y, -2z) \), indicating how the function changes in each spatial direction. This vector will help us define the tangent plane at specific points.

Partial Derivatives

Partial derivatives are key to understanding changes in multivariable functions by analyzing how a particular variable affects the outcome while keeping others constant. In the context of tangent planes, partial derivatives form the components of the gradient vector, which further help in constructing the equation of the tangent plane.

For a function like \( f(x, y, z) = x^{2} + y^{2} - z^{2} \), the partial derivatives are calculated as:

• With respect to \( x \), represented as \( \frac{\partial f}{\partial x} \), resulting in \( 2x \).
• With respect to \( y \), represented as \( \frac{\partial f}{\partial y} \), yielding \( 2y \).
• With respect to \( z \), represented as \( \frac{\partial f}{\partial z} \), with value \(-2z \).

These derivatives show the sensitivity of the function to changes in each variable independently, providing a snapshot of the surface's behavior near any given point. Through these derivatives, we can better visualize and compute the tangent planes on surfaces defined by multivariable functions.

Equation of a Plane

The equation of a plane is crucial when discussing tangent planes to surfaces in three-dimensional space. In this context, the equation is typically formed using the gradient, which acts as the normal vector. The general formula for the equation of a plane is \( ax + by + cz = d \), where \( (a, b, c) \) is the normal vector, and \( d \) is a constant.

When finding the tangent plane at a point, the gradient vector at that point provides the necessary coefficients for the plane's equation. For our given surface \( x^{2} + y^{2} - z^{2} = 0 \) at points \( (3, 4, 5) \) and \( (-4, -3, 5) \), we computed:

• \( abla f(3, 4, 5) = (6, 8, -10) \)
• \( abla f(-4, -3, 5) = (-8, -6, -10) \)

These vectors are used to write the tangent plane equations by ensuring the normal vector dot product with any point on the plane provides a consistent value corresponding to the surface shift. Therefore, the equations are:
- \( 6x + 8y - 10z = 38 \) for the plane through \( (3, 4, 5) \)
- \( -8x - 6y - 10z = -86 \) for the plane through \( (-4, -3, 5) \).
These equations represent the planes tangent to the surface at the specified points, characterized by their normal vectors obtained from the gradient.