Problem 16 - Active Calculus Multivariable

Question

Consider the function \(f\) defined by \(f(x, y)=8-x^{2}-3 y^{2}\) a. Determine \(f_{x}(x, y)\) and \(f_{y}(x, y)\). b. Find parametric equations in \(\mathbb{R}^{3}\) for the tangent line to the trace \(f(x, 1)\) at \(x=2\) c. Find parametric equations in \(\mathbb{R}^{3}\) for the tangent line to the trace \(f(2, y)\) at \(y=1\) d. State respective direction vectors for the two lines determined in (b) and \((c)\). e. Determine the equation of the plane that passes through the point \((2,1, f(2,1))\) whose normal vector is orthogonal to the direction vectors of the two lines found in (b) and (c). f. Use a graphing utility to plot both the surface \(z=8-x^{2}-3 y^{2}\) and the plane from (e) near the point \((2,1) .\) What is the relationship between the surface and the plane?

Step-by-Step Solution

Answer

In summary: a. \(f_x(x, y) = -2x\) and \(f_y(x, y) = -6y\) b. The parametric equations for the tangent line to the trace \(f(x, 1)\) at \(x=2\) are: \(x = 2 + 1t\), \(y = 1\), and \(z = 3 - 4t\). c. The parametric equations for the tangent line to the trace \(f(2, y)\) at \(y=1\) are: \(x = 2\), \(y = 1 + 1t\), and \(z = 3 - 6t\). d. Direction vector for the tangent line to the trace \(f(x, 1)\) at \(x=2\): \(\langle 1, 0, -4 \rangle\); direction vector for the tangent line to the trace \(f(2, y)\) at \(y=1\): \(\langle 0, 1, -6 \rangle\) e. The equation of the plane that passes through the point \((2,1, f(2,1))\) with normal vector orthogonal to the direction vectors of the tangent lines is: \(4x + z - 11 = 0\). f. By graphing the surface and plane near the point \((2,1)\), we can see that the tangent plane is the best linear approximation of the surface at this point.

Step 1: Find the partial derivative of f with respect to x

To find the partial derivative of f with respect to x, we will treat y as a constant and differentiate f(x, y) with respect to x: \[f_x(x, y) = \frac{\partial}{\partial x} (8 - x^2 - 3y^2) = -2x\]

Step 2: Find the partial derivative of f with respect to y

To find the partial derivative of f with respect to y, we will treat x as a constant and differentiate f(x, y) with respect to y: \[f_y(x, y) = \frac{\partial}{\partial y} (8 - x^2 - 3y^2) = -6y\] #b. Finding parametric equations for the tangent line to the trace f(x, 1) at x=2#

Step 3: Evaluate the gradient of f at (2, 1)

We have the partial derivatives, so now we can find the gradient at the given point (2,1). The gradient is given by the vector: \[\nabla f(2, 1) = \langle f_x(2,1), f_y(2,1) \rangle = \langle -2(2), -6(1) \rangle = \langle -4, -6 \rangle\]

Step 4: Find the tangent vector for the tangent line to the trace f(x, 1) at x=2

Since we are given \(x = 2\) and \(y = 1\), we can find the point on this trace as below: \[(2,1,f(2,1)) = (2, 1, 8 - 2^2 - 3(1)^2) = (2, 1, 3)\] For this trace, the tangent vector will be \(\langle 1, 0, f_x(2, 1) \rangle\), which is: \[\langle 1, 0, -4 \rangle\]

Step 5: Write the parametric equations for the tangent line to the trace f(x, 1) at x=2

Now that we have the tangent vector, we can write the parametric equations for the tangent line: \[x = 2 + 1t\] \[y = 1 + 0t\] \[z = 3 - 4t\] #c. Finding parametric equations for the tangent line to the trace f(2, y) at y=1#

Step 6: Find the tangent vector for the tangent line to the trace f(2, y) at y=1

For this trace, the tangent vector will be \(\langle 0, 1, f_y(2, 1) \rangle\), which is: \[\langle 0, 1, -6 \rangle\]

Step 7: Write the parametric equations for the tangent line to the trace f(2, y) at y=1

Now that we have the tangent vector, we can write the parametric equations for the tangent line: \[x = 2 + 0t\] \[y = 1 + 1t\] \[z = 3 - 6t\] #d. Direction vectors for the tangent lines#

Step 8: Write the direction vectors for the two tangent lines found in (b) and (c)

The direction vectors are given by: For tangent line of trace f(x, 1) at x=2: \[\langle 1, 0, -4 \rangle\] For tangent line of trace f(2, y) at y=1: \[\langle 0, 1, -6 \rangle\] #e. Finding the equation of the plane#

Step 9: Find the normal vector orthogonal to the direction vectors of the tangent lines

To find the normal vector, we need to take the cross product of the direction vectors: \[\langle 1, 0, -4 \rangle \times \langle 0, 1, -6 \rangle = \langle 4 + 0, 0 - 0, 1 - 0 \rangle = \langle 4, 0, 1 \rangle\]

Step 10: Find the equation of the plane that passes through the point (2,1, f(2,1)) with normal vector \(\langle 4, 0, 1 \rangle\)

Now that we have the normal vector and the point, we can find the equation of the plane using the point-normal form: \[4(x-2) + 0(y-1) + 1(z-3) = 0\] \[4x + z - 11 = 0\] #f. Plotting the surface and the plane to understand their relationship# By graphing the surface \(z = 8 - x^2 - 3y^2\) and the plane \(4x + z - 11 = 0\) near the point \((2,1)\), we can see that they intersect and the tangent plane is the best linear approximation of the surface at the point \((2,1)\).

Key Concepts

Partial Derivatives

Understanding partial derivatives is fundamental in multivariable calculus. A partial derivative represents the change in a multivariable function when one variable changes, while all other variables remain constant. When given a function like f(x, y), the partial derivatives are symbolically represented as \(f_x(x, y)\) and \(f_y(x, y)\) for the derivatives with respect to \(x\) and \(y\), respectively. These values are crucial as they help characterize the slope of the surface defined by the function in the direction of each variable.

In the context of the exercise, when \(f\) is defined as \(f(x, y)=8-x^2-3y^2\), computing the partial derivatives provides us with the slopes in the \(x\)-direction and \(y\)-direction at any point on the surface. The partial derivatives \(f_x\) and \(f_y\) at each point are, therefore, essential in finding tangent lines and planes.

Tangent Line

The tangent line to a curve at a given point is a straight line that touches the curve only at that point and has the same direction as the curve at that point. For functions of two variables, tangent lines can be found to the curve traces within the surface, which are curves created by holding one variable constant. These lines are practical for visualizing the slope of the function at a given point (x, y).

By utilizing partial derivatives, we can determine the tangent line to such a trace. As seen in the exercise, this involves calculating the gradient vector and utilizing it to find the tangent vector, which then allows for constructing parametric equations representing the line.

Parametric Equations

When it comes to curves in space, parametric equations offer a powerful method to describe a line's path in three dimensions. They express the coordinates of the points on a curve as functions of a parameter, noted as \(t\), which typically represents time. A set of parametric equations for a line in space will usually have the form \(x(t)\), \(y(t)\), and \(z(t)\).

In our example, the parametric equations for the tangent lines provide a clear description of the lines' directions and the specific points on the function's surface they touch. Designing the parametric equations thus consists of finding a point through which the line passes and a directional vector along which the line extends.

Gradient Vector

The gradient vector of a function, denoted as \(abla f\), points in the direction of the greatest increase of the function and its magnitude represents the rate of increase. It contains all of the function's first partial derivatives. In a geometric sense, the gradient vector at a point on a surface is perpendicular to the tangent plane to the surface at that point.

For a two-variable function such as \(f(x, y)\), the gradient vector is given by \(abla f = \langle f_x, f_y \rangle\), where \(f_x\) and \(f_y\) are the partial derivatives with respect to \(x\) and \(y\) respectively. As demonstrated in the solution to the exercise, calculating the gradient vector at a specific point is crucial to formulating the tangent line equations.

Cross Product

The cross product is a vector operation that takes two vectors in three-dimensional space and produces a third vector that is perpendicular to the plane containing the first two vectors. The length of the resulting vector is proportional to the area of the parallelogram spanned by the original vectors. This operation is denoted by the symbol \(\times\).

In the exercise, the cross product is utilized to find a normal vector to the plane defined by two direction vectors of the tangent lines. Symbolically, if \(\vec{a}\) and \(\vec{b}\) are vectors, their cross product, \(\vec{a} \times \vec{b}\), is computed using the determinant of a matrix constructed from the unit vectors \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\), along with the components of \(\vec{a}\) and \(\vec{b}\). The practical application of this process in the problem yields the normal vector needed for the equation of the tangent plane.