Problem 2 - Active Calculus Multivariable

Question

Find the equation of the tangent plane to the surface \(z=e^{2 x / 17} \ln (1 y)\) at the point (3,2,0.9865) . \(\mathrm{z}=\) __________.

Step-by-Step Solution

Answer

\[\frac{2}{17}e^{\frac{2(3)}{17}} \ln(2)(x-3) + e^{\frac{2(3)}{17}} \frac{1}{2}(y-2) = 0\]

Step 1: Compute the partial derivatives

Firstly, we need to find the partial derivatives of the given function with respect to x and y. Our function is: \[z = e^{(2x / 17)} \ln(y)\] So, the partial derivative with respect to x is: \[\frac{\partial z}{\partial x} = \frac{2}{17}e^{(2x / 17)} \ln(y)\] And the partial derivative with respect to y is: \[\frac{\partial z}{\partial y} = e^{(2x / 17)} \frac{1}{y}\]

Step 2: Calculate the gradient vector at the given point

Now, we need to evaluate the partial derivatives at the point (3, 2, 0.9865). \[\frac{\partial z}{\partial x}(3,2) = \frac{2}{17}e^{(2(3) / 17)} \ln(2)\] \[\frac{\partial z}{\partial y}(3,2) = e^{(2(3) / 17)} \frac{1}{2}\] These two values will form the gradient vector, which represents the normal vector to the tangent plane. The gradient vector is \(\begin{pmatrix}\frac{2}{17}e^{(2(3) / 17)} \ln(2) \\ e^{(2(3) / 17)} \frac{1}{2}\end{pmatrix}\).

Step 3: Use the point-normal form of the plane's equation

Now that we have the normal vector to the tangent plane, we can use the point-normal form of the plane's equation to find the equation of the tangent plane. The point-normal form of the plane's equation is: \[A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\] Given that the point is (3, 2, 0.9865) and the normal vector components are A = \(\frac{2}{17}e^{(2(3) / 17)} \ln(2)\) and B = \(e^{(2(3) / 17)} \frac{1}{2}\), the equation of the tangent plane is: \[\frac{2}{17}e^{(2(3) / 17)} \ln(2)(x-3) + e^{(2(3) / 17)} \frac{1}{2}(y-2) + C(z-0.9865) = 0\]

Step 4: Solve for C of the normal vector

To find the value of C, we know that the tangent plane will be perpendicular to the gradient vector, which can be computed using the dot product of the gradient: \[C_1\frac{\partial z}{\partial x} +C_2\frac{\partial z}{\partial y} + C(z-z_0) = 0\] with C_1= \(\frac{2}{17}e^{(2(3) / 17)} \ln(2)\) and C_2=\(e^{(2(3) / 17)} \frac{1}{2}\) To find C, substitute x = 3, y = 2, and z = 0.9865: \[C \cdot 0.9865 = 0\] Either C = 0, or we already plugged the tangent plane point into our partials, so it's redundant. Thus, we don't need to consider C in our equation.

Step 5: Write the final equation of the tangent plane

Now we can plug all these values into our formula to find the equation of the tangent plane: \[\frac{2}{17}e^{(2(3) / 17)} \ln(2)(x-3) + e^{(2(3) / 17)} \frac{1}{2}(y-2) = 0\] This is the final equation of the tangent plane to the surface z = e^(2x/17)ln(1y) at the point (3, 2, 0.9865).

Key Concepts

Partial Derivatives

Partial derivatives are crucial in finding the tangent plane to a surface. They help determine how a function changes in specific directions. Imagine it as the slope of the surface if you move only along the x-axis or the y-axis. For a function defined by two variables, let's say a surface given by \(z = f(x, y)\), we can compute:

• The partial derivative with respect to \(x\), denoted as \(\frac{\partial z}{\partial x}\)
• The partial derivative with respect to \(y\), denoted as \(\frac{\partial z}{\partial y}\)

These derivatives give us the rate of change of the function in each direction, helping us determine how steep the surface is in those particular directions. This information is the foundation of constructing the tangent plane equation.

Gradient Vector

The gradient vector is a vector that contains all the partial derivatives of a function and provides the direction of the steepest increase of the function. For our surface \(z = f(x, y)\), the gradient vector can be written as: \[abla z = \begin{pmatrix}\frac{\partial z}{\partial x} \ \frac{\partial z}{\partial y}\end{pmatrix}\] Imagine the gradient vector as a compass that points towards the steepest path up a hill. Therefore, it is directly perpendicular to the tangent plane of the surface. Evaluating this vector at a specific point gives us the normal vector, which is essential for forming the equation of the tangent plane.

Point-Normal Form

The point-normal form of a plane's equation is a method used to find the equation of a plane when a normal vector and a point on the plane are known. The general form is: \[A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\] Here, \((x_0, y_0, z_0)\) is the point on the plane, and \(A, B,\) and \(C\) represent the components of the normal vector. The beauty of this form is that it directly uses the normal vector's components, making it a straightforward approach to formulating the tangent plane equation. The formula ensures that the plane is orthogonal to the normal vector, neatly fitting in a 3D mathematical space.

Normal Vector

The normal vector plays a pivotal role when determining the tangent plane. It is a vector that is perpendicular to the plane, effectively defining its orientation in space. In our context, the normal vector is obtained from the gradient vector. So, when we find the gradient vector of our surface's equation, we effectively have our normal vector.

• The tangent plane is orthogonal to this normal vector.
• Having this vector allows us to formulate the point-normal form equation, necessary for the plane's representation.

The importance of the normal vector is that it ensures that the tangent plane accurately represents the surface's local linear approximation at a given point.