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

(a) \(x=2+t, y=1, z=9+12t\); (b) \(x=2, y=1+t, z=9+10t\); (c) \(x=2+t, y=1+t, z=9+22t\).

Step 1: Understand the Surface and Tangent Line Problem

The surface is given by the equation \(z = y^2 + x^3 y\). We are to find parametric equations for a tangent line at the point \((2, 1, 9)\) having specific orientations in the \(xy\)-plane.

Step 2: Calculate Partial Derivatives for Tangent Line

To find the equations of the tangent line, we need to calculate the partial derivatives of the surface \(z = y^2 + x^3 y\):\ \(\frac{\partial z}{\partial x} = 3x^2 y\); \(\frac{\partial z}{\partial y} = 2y + x^3\). Evaluate these at \((x, y) = (2, 1)\):\[\frac{\partial z}{\partial x} \Big|_{(2,1)} = 3(2)^2(1) = 12\]\[\frac{\partial z}{\partial y} \Big|_{(2,1)} = 2(1) + (2)^3 = 2 + 8 = 10 \]

Step 3: Establish Vector Form for Tangent Line

The tangent vector can be written as \( (dx, dy, dz) = (a, b, 12a + 10b) \) where \(a\) and \(b\) are parameters based on the different requirements for projection in the \(xy\)-plane.

Step 4: (a) Projection Parallel to the x-axis

For the projection to be parallel to the \(x\)-axis, we set \(b = 0\). Therefore, the tangent vector becomes \((a, 0, 12a)\). Choose \(a = 1\) for simplicity, leading to the parametric equations:\[ x = 2 + t, \, y = 1, \, z = 9 + 12t \]

Step 5: (b) Projection Parallel to the y-axis

For the projection to be parallel to the \(y\)-axis, set \(a = 0\). Thus, the tangent vector simplifies to \((0, b, 10b)\). Choosing \(b = 1\), the parametric equations become:\[ x = 2, \, y = 1 + t, \, z = 9 + 10t \]

Step 6: (c) Projection Parallel to Line x = y

Here, the projection condition \(x = y\) implies \(a = b\). The tangent vector becomes \((a, a, 12a + 10a) = (a, a, 22a)\). Choose \(a = 1\) to find the parametric equations:\[ x = 2 + t, \, y = 1 + t, \, z = 9 + 22t \]

Key Concepts

Parametric Equations

Parametric equations are a method of expressing geometric objects using a parameter, usually denoted by the letter \(t\). This approach helps in describing paths and curves in space. Each coordinate of a point is expressed as a separate function of this parameter. For example, the line in three-dimensional space passing through a point can be described using three parametric equations.

In our problem, the goal is to find a set of parametric equations for a line tangent to a surface. At the point \(2, 1, 9\), the tangent line has different orientations in the \(xy\)-plane. To achieve this, we identify parametric forms parallel to the \(x\)-axis, \(y\)-axis, and the line \(x = y\). By setting unique conditions for each axis, we use parameters that control the proportion of movement in each direction.

• For projection parallel to the \(x\)-axis, we use \(x = 2 + t\), showcasing movement in the x-direction while keeping \(y\) constant.
• For projection parallel to the \(y\)-axis, we have \(y = 1 + t\), highlighting constant \(x\) as \(y\) changes.
• For \(x = y\), the parameter sets both \(x\) and \(y\) to change by the same amount, \(x = 2 + t\) and \(y = 1 + t\).

Tangent Line

A tangent line to a surface at a given point is a line that touches the surface at that point and runs straight through it without curving. This line is an extension of the tangent plane, which barely skims the surface.

For a surface expressed in the equation \(z = f(x, y)\), the tangent line at a point summarises the rate of change in the \(z\)-direction concerning movements in the \(x\) and \(y\) directions.

• The tangent line incorporates a direction vector derived from the surface's partial derivatives.
• In our exercise, we derived the line by taking partial derivatives of \(z\) to construct vector components.
• Each scenario modifies these components to meet specific requirements for orientation in the \(xy\)-plane.

Thus, the tangent line acts as a straight path along the surface where the point \( (2, 1, 9) \) resides.

Partial Derivatives

Partial derivatives are essential in understanding how a multivariable function changes as its variables change. For a function \(z = f(x, y)\), partial derivatives bring insight into how \(z\) changes separately with \(x\) and \(y\).

These derivatives are crucial in finding the tangent line to a surface because they unveil the slope of the surface in various directions.

• The partial derivative \(\frac{\partial z}{\partial x}\) reveals the slope concerning the x-axis.
• The partial derivative \(\frac{\partial z}{\partial y}\) indicates the slope concerning the y-axis.

In the given problem, these derivatives were calculated at the point \( (2, 1) \), resulting in values of \(12\) and \(10\), respectively. These values serve as weights determining how changes in \(x\) and \(y\) influence changes in \(z\). This leads to crafting the equation for the tangent line efficiently.

Surface Projection

Surface projection is the method of viewing a 3D surface as a shadow or outline on a 2D plane, such as the \(xy\)-plane. By projecting the tangent line onto this plane, we can visualize its specific orientation and alignment.

In this task, projections determined how the tangent line behaves concerning the axes:

• Parallel to the \(x\)-axis implies a change in \(x\) with no change in \(y\).
• Parallel to the \(y\)-axis means a variation in \(y\) while \(x\) remains constant.
• Aligned with the line \(x = y\) expresses equal changes in both \(x\) and \(y\).

By adjusting parameters, the line's projection can shift as required, thereby illustrating the geometry and motion of the surface in the \(xy\)-plane.