Problem 42 - Calculus Early Transcendentals

Question

Suppose you need to know an equation of the tangent plane to a surface \(S\) at the point \(P(2,1,3) .\) You don't have an equation for \(S\) but you know that the curves $$\begin{aligned} \mathbf{r}_{1}(t) &=\left\langle 2+3 t, 1-t^{2}, 3-4 t+t^{2}\right\rangle \\ \mathbf{r}_{2}(u) &=\left\langle 1+u^{2}, 2 u^{3}-1,2 u+1\right\rangle \end{aligned}$$ both lie on \(S .\) Find an equation of the tangent plane at \(P .\)

Step-by-Step Solution

Answer

The tangent plane equation is \( 24x - 14y + 18z = 68 \).

Step 1: Understand the Problem

You need to find the tangent plane to a surface \( S \) at a point \( P(2,1,3) \). Two vector-valued functions \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(u) \) are given that lie on \( S \). To find the tangent plane, you need two tangent vectors at point \( P \) on the curves and then use these to find the normal vector to the plane.

Step 2: Calculate Tangent Vector from \( \mathbf{r}_1(t) \)

The position vector \( \mathbf{r}_1(t) \) is given as \( \left\langle 2+3t, 1-t^2, 3-4t+t^2 \right\rangle \). The tangent vector is found by taking the derivative of \( \mathbf{r}_1(t) \) with respect to \( t \):\[ \mathbf{r}_1'(t) = \left\langle 3, -2t, -4 + 2t \right\rangle \].Evaluate this at \( t = 0 \) since \( \mathbf{r}_1(0) = (2,1,3) \):\[ \mathbf{r}_1'(0) = \left\langle 3, 0, -4 \right\rangle \].

Step 3: Calculate Tangent Vector from \( \mathbf{r}_2(u) \)

The position vector \( \mathbf{r}_2(u) \) is given as \( \left\langle 1+u^2, 2u^3-1, 2u+1 \right\rangle \). The tangent vector is found by taking the derivative of \( \mathbf{r}_2(u) \) with respect to \( u \):\[ \mathbf{r}_2'(u) = \left\langle 2u, 6u^2, 2 \right\rangle \].Evaluate this at \( u = 1 \) since \( \mathbf{r}_2(1) = (2,1,3) \):\[ \mathbf{r}_2'(1) = \left\langle 2, 6, 2 \right\rangle \].

Step 4: Calculate Normal Vector to the Plane

The normal vector to the plane can be found by taking the cross product of the tangent vectors \( \mathbf{r}_1'(0) = \left\langle 3, 0, -4 \right\rangle \) and \( \mathbf{r}_2'(1) = \left\langle 2, 6, 2 \right\rangle \):\[ \mathbf{n} = \mathbf{r}_1'(0) \times \mathbf{r}_2'(1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 0 & -4 \ 2 & 6 & 2 \end{vmatrix} \].Calculate this determinant:\[ \mathbf{n} = \left\langle 0 \cdot 2 - (-4) \cdot 6, - (3 \cdot 2 - (-4) \cdot 2), 3 \cdot 6 - 0 \cdot 2 \right\rangle = \left\langle 24, -14, 18 \right\rangle \].

Step 5: Write the Equation of the Plane

The equation of a plane with normal vector \( \mathbf{n} = \left\langle a, b, c \right\rangle \) passing through point \( (x_0, y_0, z_0) \) is: \[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \].Substitute \( a = 24 \), \( b = -14 \), \( c = 18 \), and the point \( (2,1,3) \):\[ 24(x-2) - 14(y-1) + 18(z-3) = 0 \].Simplify the equation:\[ 24x - 14y + 18z = 68 \].

Step 6: Conclusion

The equation of the tangent plane to the surface at point \( P(2, 1, 3) \) is given by:\[ 24x - 14y + 18z = 68 \].

Key Concepts

vector-valued functions

Vector-valued functions play a significant role in understanding curves and surfaces in three-dimensional space. Essentially, these functions map real numbers to vectors. For example, given a parameter, they provide a position vector that traces a path in the vector space. It means that as you input different values, you get a set of vectors indicating various points along a curve.

In our problem, we have two vector-valued functions \( \mathbf{r}_1(t) = \langle 2 + 3t, 1 - t^2, 3 - 4t + t^2 \rangle \) and \( \mathbf{r}_2(u) = \langle 1 + u^2, 2u^3 - 1, 2u + 1 \rangle \). These functions lie on the surface \( S \) and describe the curves on that surface passing through a specific point. Understanding these functions is crucial because they provide the necessary paths on a surface to help us determine the plane's direction at any given point.

normal vector

The normal vector is vital when defining the orientation of a plane. It is a vector that is perpendicular to every vector lying on that plane. When you have the normal vector, you can easily find the tangent plane's equation since it directly reflects the plane's inclination in three-dimensional space.

For our exercise, the aim is to find the normal vector to the tangent plane at a given point. After calculating tangent vectors from the given vector-valued functions, we use the cross product to get the normal vector. This process ensures the obtained normal vector is perpendicular to the tandem of tangent vectors on the plane.

tangent vectors

Tangent vectors describe the direction and rate of change of a curve at a particular point. They indicate how a curve progresses from one point to the next. In mathematical terms, tangent vectors result from differentiating a vector-valued function with respect to its parameter.

For instance, in our exercise, we calculate the derivative of \( \mathbf{r}_1(t) \) to find \( \mathbf{r}_1'(t) = \langle 3, -2t, -4 + 2t \rangle \) and evaluate it at \( t = 0 \), obtaining \( \mathbf{r}_1'(0) = \langle 3, 0, -4 \rangle \). Similarly, for \( \mathbf{r}_2(u) \), we derive \( \mathbf{r}_2'(u) = \langle 2u, 6u^2, 2 \rangle \) and evaluate at \( u = 1 \), giving us \( \mathbf{r}_2'(1) = \langle 2, 6, 2 \rangle \). These vectors help define the plane's surface tangent at a specific point.

cross product

The cross product is a fundamental operation in vector algebra, used to find a vector perpendicular to two given vectors. Geometrically, if you have two vectors describing curves on a surface, their cross product provides a way to find a normal vector to the tangent plane.

For the problem at hand, the cross product of the tangent vectors \( \mathbf{r}_1'(0) = \langle 3, 0, -4 \rangle \) and \( \mathbf{r}_2'(1) = \langle 2, 6, 2 \rangle \) is computed as:

• \( \mathbf{n} = \left\langle 0 \cdot 2 - (-4) \cdot 6, - (3 \cdot 2 - (-4) \cdot 2), 3 \cdot 6 - 0 \cdot 2 \right\rangle = \left\langle 24, -14, 18 \right\rangle \).

This resulting vector \( \mathbf{n} = \langle 24, -14, 18 \rangle \) is perpendicular to both original vectors, making it an ideal normal vector for constructing the tangent plane equation.