Problem 21 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: 1. At point (2,2,4): 12(x-2) - 4(y-2) - (z-4) = 0 2. At point (-1,-1,1): -e^(-2)(y+1) - (z-1) = 0

Step 1: Compute partial derivatives

To find the tangent plane, we need the partial derivatives of the function z with respect to x and y. The function z = x^2 * e^(x-y), so we can find its partial derivatives as follows: $$ \frac{\partial z}{\partial x} = 2x \cdot e^{x-y} + x^2 \cdot e^{x-y} $$ and $$ \frac{\partial z}{\partial y} = -x^2 \cdot e^{x-y} $$ Now, we will use these partial derivatives along with the given coordinates to find the tangent plane equation at each of the points:

Step 2: Find the normal vector at (2,2,4)

We calculate the partial derivatives at the point (2,2,4). Plugging in the values of x=2, y=2 into the equations, we get: $$ \frac{\partial z}{\partial x}\bigg{|}_{(2,2,4)} = 2(2) e^{2-2} + (2)^2 e^{2-2} = 12 $$ and $$ \frac{\partial z}{\partial y}\bigg{|}_{(2,2,4)} = -(2)^2 e^{2-2} = -4 $$ Thus, the normal vector at point (2,2,4) is (12, -4, -1).

Step 3: Write the tangent plane equation at (2,2,4)

Let (x, y, z) be a general point on the tangent plane. Then, the normal vector (12, -4, -1) will be orthogonal to the vector (x-2, y-2, z-4). Therefore, we get the following equation for the tangent plane at (2,2,4): $$ (12)(x-2) + (-4)(y-2) + (-1)(z-4) = 0 $$ Now, we will do the same steps for the point (-1,-1,1).

Step 4: Find the normal vector at (-1,-1,1)

We calculate the partial derivatives at the point (-1,-1,1). Plugging in the values of x=-1, y=-1 into the equations, we get: $$ \frac{\partial z}{\partial x}\bigg{|}_{(-1,-1,1)} = 2(-1) e^{-1-1} + (-1)^2 e^{-1-1} = -e^{-2} + e^{-2} = 0 $$ and $$ \frac{\partial z}{\partial y}\bigg{|}_{(-1,-1,1)} = -(-1)^2 e^{-1-1} = -e^{-2} $$ Thus, the normal vector at point (-1,-1,1) is (0, -e^{-2}, -1).

Step 5: Write the tangent plane equation at (-1,-1,1)

Let (x, y, z) be a general point on the tangent plane. Then, the normal vector (0, -e^{-2}, -1) will be orthogonal to the vector (x+1, y+1, z-1). Therefore, we get the following equation for the tangent plane at (-1,-1,1): $$ (0)(x+1) + (-e^{-2})(y+1) + (-1)(z-1) = 0 $$ For the two given points, the equations of the tangent planes are: 1. At point (2,2,4): $$ 12(x-2) - 4(y-2) - (z-4) = 0 $$ 2. At point (-1,-1,1): $$ -e^{-2}(y+1) - (z-1) = 0 $$

Key Concepts

Partial Derivatives

Partial derivatives are essential in finding tangent planes to surfaces. They measure how a function changes as its input variables change. If a function has multiple variables, partial derivatives are the derivatives taken with respect to one variable, keeping others constant.

For the function \(z = x^{2} e^{x-y}\), we compute the partial derivatives with respect to \(x\) and \(y\):

• \(\frac{\partial z}{\partial x} = 2x \cdot e^{x-y} + x^{2} \cdot e^{x-y}\)
• \(\frac{\partial z}{\partial y} = -x^{2} \cdot e^{x-y}\)

These derivatives give the rate of change of \(z\) with respect to \(x\) and \(y\), respectively. They are crucial in determining how the surface behaves at different points. By evaluating these partial derivatives at specific points, we can gain information about the slope or incline of the surface in the direction of each variable.

Normal Vector

The normal vector of a surface is perpendicular to the tangent plane at a given point. It is key to finding the equation of the tangent plane. In mathematical terms, it describes the direction in which the surface climbs steepest. To find the normal vector at a point \((x_0, y_0, z_0)\) on the surface, we use the partial derivatives.

For the point \((2,2,4)\), we computed:

• \(\frac{\partial z}{\partial x} = 12\)
• \(\frac{\partial z}{\partial y} = -4\)
• Together with the derivative with respect to \(z\), which is always \(-1\), form the normal vector: \((12, -4, -1)\)

Analogously, the normal vector at \((-1,-1,1)\) is derived from:

• \(\frac{\partial z}{\partial x} = 0\)
• \(\frac{\partial z}{\partial y} = -e^{-2}\)
• Resulting in the normal vector: \((0, -e^{-2}, -1)\)

These vectors help construct the tangent plane equation by informing us about the slope in all three-dimensional directions.

Surface Equation

The equation of the tangent plane provides a linear approximation to the surface at a given point. It is constructed using the normal vector and a point on the surface. The general formula for a tangent plane to a surface \(z = f(x, y)\) at \((x_0, y_0, z_0)\) is:

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

Here, \((a, b, c)\) is the normal vector.

Let's consider the point \((2,2,4)\):

• Using the normal vector \((12, -4, -1)\), the equation is: \[12(x-2) - 4(y-2) - (z-4) = 0 \]

At the point \((-1,-1,1)\):

• Using the normal vector \((0, -e^{-2}, -1)\), the equation is: \[-e^{-2}(y+1) - (z-1) = 0 \]

These tangent plane equations provide a geometric insight into the surface's behavior near the points by showing a flat approximation of the surface at those specific points.