Problem 16 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$2 x+y^{2}-z^{2}=0 ;(0,1,1) \text { and }(4,1,-3)$$

Step-by-Step Solution

Answer

Question: Find the equations of the tangent planes to the surface given by the equation \(2x+y^2-z^2=0\) at the points \((0,1,1)\) and \((4,1,-3)\). Answer: The tangent planes at the given points on the surface are \(2x+2y-2z=0\) and \(2x+2y+6z=-2\).

Step 1: Compute the Gradient of the Surface Equation

The given surface equation is: $$2x + y^2 - z^2 = 0$$ To find the gradient, we will take the partial derivatives with respect to each variable: $$\begin{aligned} \frac{\partial f}{\partial x} &= 2,\\ \frac{\partial f}{\partial y} &= 2y,\\ \frac{\partial f}{\partial z} &= -2z. \end{aligned}$$ This gives us the gradient vector: $$\nabla f = \langle 2, 2y, -2z \rangle$$

Step 2: Find the Normal Vectors at Each Point

Now plug the coordinates of each point \((0,1,1)\) and \((4,1,-3)\) into the gradient vector to find the normal vectors of the tangent planes at each point. For point \((0,1,1)\): $$\nabla f(0,1,1) = \langle 2, 2(1), -2(1) \rangle = \langle 2, 2, -2 \rangle = \mathbf{n_1}$$ For point \((4,1,-3)\): $$\nabla f(4,1,-3) = \langle 2, 2(1), -2(-3) \rangle = \langle 2, 2, 6 \rangle = \mathbf{n_2}$$

Step 3: Use the Point-Normal Form to Find the Plane Equations

The point-normal form of a plane equation is given by \((\mathbf{n} \cdot \mathbf{v})=0\), where \(\mathbf{n}\) is the normal vector and \(\mathbf{v}=\langle x-x_0, y-y_0, z-z_0 \rangle\) is a position vector pointing from a point on the plane \((x_0, y_0, z_0)\) to a general point \((x, y, z)\). For the first tangent plane, with point \((0,1,1)\) and normal vector \(\mathbf{n_1} = \langle 2, 2, -2 \rangle\): $$\begin{aligned} (\mathbf{n_1} \cdot \mathbf{v})&= \langle 2, 2, -2 \rangle \cdot \langle x-0, y-1, z-1 \rangle \\ &= 2(x-0)+2(y-1)-2(z-1)=0 \\ \Rightarrow 2x + 2(y-1) - 2(z-1) &= 0 \\ \Rightarrow 2x+2y-2z&=0 \end{aligned}$$ Hence, the equation of the first tangent plane is: $$2x+2y-2z=0$$ For the second tangent plane, with point \((4,1,-3)\) and normal vector \(\mathbf{n_2} = \langle 2, 2, 6 \rangle\): $$\begin{aligned} (\mathbf{n_2} \cdot \mathbf{v})&= \langle 2, 2, 6 \rangle \cdot \langle x-4, y-1, z+3 \rangle \\ &= 2(x-4)+2(y-1)+6(z+3)=0 \\ \Rightarrow 2(x-4) + 2(y-1) + 6(z+3) &= 0 \\ \Rightarrow 2x+2y+6z&=-2 \end{aligned}$$ Hence, the equation of the second tangent plane is: $$2x+2y+6z=-2$$ So, the tangent planes at the given points on the surface are: $$2x+2y-2z=0$$ and $$2x+2y+6z=-2$$

Key Concepts

Gradient Vector

The gradient vector is a fundamental concept in vector calculus. When working with functions of several variables, the gradient vector, often denoted as \(abla f\), provides the direction of the steepest ascent of the function. This makes it particularly useful for finding tangent planes to surfaces.

To compute the gradient vector, we need to take the partial derivatives of the function with respect to each of its variables. In our problem, the surface is described by the equation \(2x + y^2 - z^2 = 0\). By taking the partial derivatives:

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

This results in the gradient vector \(abla f = \langle 2, 2y, -2z \rangle\). The gradient delivers both direction and magnitude that are crucial in determining tangent planes.

Point-Normal Form

In geometry, the point-normal form is a valuable tool for defining planes using vectors. The equation of a plane can be found if we know one point on the plane and the normal vector, which is perpendicular to the plane.

For a point \((x_0, y_0, z_0)\) and a normal vector \(\mathbf{n} = \langle a, b, c \rangle\), the point-normal form of a plane is:

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

This equation assures that for any point \((x, y, z)\) on the plane, the vector from \((x_0, y_0, z_0)\) to \((x, y, z)\) is perpendicular to \(\mathbf{n}\).

In our exercise, for point \((0,1,1)\) with normal vector \(\langle 2, 2, -2 \rangle\), the tangent plane's equation becomes \(2(x-0) + 2(y-1) - 2(z-1) = 0\), simplifying to \(2x + 2y - 2z = 0\). This use of the normal vector is crucial in translating geometric intuition into algebraic form.

Partial Derivatives

Partial derivatives represent how a function changes with respect to one variable while keeping other variables constant. This concept is essential in multivariable calculus for understanding how surfaces behave.

In our problem, we have the function \(f(x, y, z) = 2x + y^2 - z^2\), and we computed its partial derivatives to form the gradient. For instance:

• \(\frac{\partial f}{\partial x}\) signifies the rate of change of \(f\) as \(x\) changes, keeping \(y\) and \(z\) constant, calculated as \(2\).
• \(\frac{\partial f}{\partial y}\) represents the change in \(f\) with respect to \(y\), holding \(x\) and \(z\) steady, resulting in \(2y\).
• \(\frac{\partial f}{\partial z}\) shows how \(f\) varies with \(z\), with all else constant, given by \(-2z\).

Understanding these partial derivatives is crucial because they comprise the gradient vector, which is used to find the tangent plane. This knowledge links function behavior to physical surfaces.