Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find parametric equations of the line tangent to the surface \(z=y^{2}+x^{3} y\) at the point \((2,1,9)\) whose projection on the \(x y\) -plane is (a) parallel to the \(x\) -axis; (b) parallel to the \(y\) -axis; (c) parallel to the line \(x=y\).

Step-by-Step Solution

Answer

The parametric equations are: (a) parallel to x-axis, \(x=2+t, y=1, z=9+12t\); (b) parallel to y-axis, \(x=2, y=1+t, z=9+10t\); (c) parallel to line \(x=y\), \(x=2+t, y=1+t, z=9+22t\).

Step 1: Find Partial Derivatives

First, find the partial derivatives of the surface equation \(z = y^2 + x^3 y \) with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \(f_x = 3x^2 y\), and with respect to \(y\) is \(f_y = 2y + x^3\). This gives the direction of the tangent plane at any point.

Step 2: Calculate at Point (2,1)

Evaluate the partial derivatives found in Step 1 at the point \((2, 1)\). For \(f_x\), substitute \(x = 2\) and \(y = 1\) to get \(f_x(2,1) = 3(2)^2(1) = 12\). For \(f_y\), substitute \(x = 2\) and \(y = 1\) to get \(f_y(2,1) = 2(1) + 2^3 = 2 + 8 = 10\). Hence, the gradient at the point is \((12, 10)\).

Step 3: Find Tangent Line with Projection Parallel to x-axis

For the tangent line with projection parallel to the \(x\)-axis, the direction vector in the \(xy\)-plane should have a \(y\)-component of 0. Therefore, the direction vector is \(\mathbf{v_1} = (1, 0)\). The associated parametric equations are \(x = 2 + t\), \(y = 1\), \(z = 9 + 12t\).

Step 4: Find Tangent Line with Projection Parallel to y-axis

For the tangent line with projection parallel to the \(y\)-axis, the direction vector should have an \(x\)-component of 0. Therefore, the direction vector is \(\mathbf{v_2} = (0, 1)\). The associated parametric equations are \(x = 2\), \(y = 1 + t\), \(z = 9 + 10t\).

Step 5: Find Tangent Line with Projection Parallel to Line x=y

For the tangent line whose projection is parallel to the line \(x=y\), the direction vector should be \(\mathbf{v_3} = (1, 1)\). The associated parametric equations are \(x = 2 + t\), \(y = 1 + t\), \(z = 9 + 22t\), using the sum \(12 + 10 = 22\) for the \(z\)-component since \(f_x + f_y = 12 + 10\).

Key Concepts

Parametric Equations

When dealing with curves and lines, parametric equations are a useful tool. They allow us to express coordinates as functions of a parameter, typically denoted as \( t \). This method is favored because it provides a clear way to describe the motion of a point along a curve or line.

In the given problem, parametric equations are used to find lines tangent to a surface at a specific point. These equations consist of three separate formulas, each for one of the coordinates \( x \), \( y \), and \( z \). An example from the solution is when the projection is parallel to the \( x \)-axis:

• \( x = 2 + t \)
• \( y = 1 \)
• \( z = 9 + 12t \)

Each equation shows how the coordinates change as the parameter \( t \) changes, starting from the point \((2, 1, 9)\). This method provides a structured approach to finding tangent lines in three dimensions.

Tangent Line

A tangent line to a surface at a point is a line that just "touches" the surface at that point without "cutting" through it. To understand this better, imagine pressing a pencil against a ball—the pencil line on the ball is similar to a tangent line.

In mathematics, finding a tangent line involves determining both the direction and position of the line. The direction is often conveyed through a vector that describes how the line "travels" through space. For instance, if a line is parallel to the \( x \)-axis, the direction vector might be \((1,0,0)\).

The solution to the exercise included finding tangent lines with specified projections. These directions relate to how the line would appear "flattened" onto the \( xy \)-plane. Using this perspective, we solve for lines whose projections were parallel to the \( x \)-axis, \( y \)-axis, and the line \( x = y \). Each projection gives the line a distinct direction before calculating its precise position.

Partial Derivatives

Partial derivatives provide a way to determine the rate of change of a function with respect to one of its variables, holding the others constant. This is particularly useful in functions with multiple variables, like the surface equation given: \( z = y^2 + x^3 y \).

In solving for the tangent line, the partial derivatives \( f_x \) and \( f_y \) were calculated to identify the changes in the surface regarding \( x \) and \( y \), respectively. These derivatives helped us determine the slope of the tangent plane:

• \( f_x = 3x^2 y \)
• \( f_y = 2y + x^3 \)

By evaluating these partial derivatives at the point \((2, 1)\), we obtained values \( f_x = 12 \) and \( f_y = 10 \). This led to identifying a direction vector for the plane, which plays a crucial role when projecting onto different axes.

Gradient Vector

The gradient vector is a critical concept in multivariable calculus, representing the slope of a surface. For a function \( f(x, y) \), the gradient is a vector containing all its partial derivatives: \( abla f = (f_x, f_y) \).

In this exercise, the gradient vector at the point \((2, 1)\) was \((12, 10)\). This vector gives the direction of the steepest ascent on the surface from that point. Furthermore, it's instrumental in determining tangent planes and lines, as it reflects the surface's incline. With the gradient known, it becomes easier to determine how the surface "tilts" at any specific point.

This knowledge is essential when projecting the tangent line's direction onto the \( xy \)-plane, ensuring that the line accurately follows the surface's curvature. Understanding the gradient equips us to handle problems involving slopes and inclinations efficiently.