Problem 24 - APEX Calculus

Question

An implicitly defined function of \(x, y\) and \(z\) is given along with a point \(P\) that lies on the surface. Use the gradient \(\nabla F\) to: (a) find the equation of the normal line to the surface at \(P,\) and (b) find the equation of the plane tangent to the surface at \(P\). \(\sin (x y)+\cos (y z)=0,\) at \(P=(2, \pi / 12,4)\)

Step-by-Step Solution

Answer

Normal line: \( x = 2 + \frac{\pi \sqrt{3}}{24}t, y = \frac{\pi}{12} - \frac{3\sqrt{3}}{2}t, z = 4 - \frac{\pi \sqrt{3}}{24}t \). Tangent plane: \( \frac{\pi \sqrt{3}}{24}(x - 2) - \frac{3\sqrt{3}}{2}(y - \frac{\pi}{12}) - \frac{\pi \sqrt{3}}{24}(z - 4) = 0 \).

Step 1: Understand the Function

The implicit function given is \( \sin(xy) + \cos(yz) = 0 \). We need to evaluate this function at the point \( P = (2, \pi/12, 4) \). Note that the function defines a surface in three-dimensional space.

Step 2: Find Partial Derivatives

Calculate the partial derivatives of the function:- \( \frac{\partial F}{\partial x} = y \cos(xy) \).- \( \frac{\partial F}{\partial y} = x \cos(xy) - z \sin(yz) \).- \( \frac{\partial F}{\partial z} = -y \sin(yz) \).

Step 3: Evaluate the Gradient at P

Substitute \( P = (2, \pi/12, 4) \) into the partial derivatives to find the gradient vector:- \( \frac{\partial F}{\partial x} = \frac{\pi}{12} \cos\left(\frac{\pi}{6}\right) = \frac{\pi \sqrt{3}}{24} \).- \( \frac{\partial F}{\partial y} = 2 \cos\left(\frac{\pi}{6}\right) - 4 \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - 2\sqrt{3} = -\frac{3\sqrt{3}}{2} \).- \( \frac{\partial F}{\partial z} = -\frac{\pi}{12} \sin\left(\frac{\pi}{3}\right) = -\frac{\pi \sqrt{3}}{24} \).Thus, \( abla F = \left( \frac{\pi \sqrt{3}}{24}, -\frac{3\sqrt{3}}{2}, -\frac{\pi \sqrt{3}}{24} \right) \).

Step 4: Equation of the Normal Line

The normal line at \( P \) is given by using the gradient vector as the direction vector:\[ x = 2 + \frac{\pi \sqrt{3}}{24}t, \quad y = \frac{\pi}{12} - \frac{3\sqrt{3}}{2}t, \quad z = 4 - \frac{\pi \sqrt{3}}{24}t \] for parameter \( t \).

Step 5: Equation of the Tangent Plane

The tangent plane at \( P \) can be found using the gradient:\[ \frac{\pi \sqrt{3}}{24}(x - 2) - \frac{3\sqrt{3}}{2}(y - \frac{\pi}{12}) - \frac{\pi \sqrt{3}}{24}(z - 4) = 0 \]This equation represents the tangent plane to the surface at point \( P \).

Key Concepts

Gradients

A gradient is like a compass that points in the direction of greatest increase of a function. For a function of several variables like our implicit function, the gradient is a vector. This vector consists of all the partial derivatives of the function.
The gradient \(abla F\) is super important because it tells us how our function is changing in every direction at a point.
When we have our gradient calculated, it helps us understand the surface we're dealing with, especially at a point P given in the problem.

• In our exercise, the gradient is \( \left( \frac{\pi \sqrt{3}}{24}, -\frac{3\sqrt{3}}{2}, -\frac{\pi \sqrt{3}}{24} \right) \).
• This vector is normal to the surface at that point, which is why it's so useful in finding normal lines and tangent planes.

Always remember: gradients point to higher ground, guiding you in navigating multivariable landscapes.

Partial Derivatives

Partial derivatives measure how a function changes as each variable changes individually. Imagine holding one variable constant and seeing how the function behaves as you vary just another.
They are the building blocks of the gradient vector, where each entry corresponds to the rate of change with respect to one variable.
In the problem:

• \( \frac{\partial F}{\partial x} = y \cos(xy) \)
• \( \frac{\partial F}{\partial y} = x \cos(xy) - z \sin(yz) \)
• \( \frac{\partial F}{\partial z} = -y \sin(yz) \)

Each of these gives us a glimpse of the slope of the surface in the respective directions (x, y, z) at any given point.
Once we substitute our specific point P into these expressions, we obtain the values needed to construct the gradient.

Tangent Planes

A tangent plane to a surface at a given point is a plane that just grazes the surface at that point. It's like a flat piece of paper touching a balloon at just one point and having the same slope at that contact point.
The equation for this tangent plane is derived using the gradient vector. The gradient at point P gives the normal vector to our surface there.

• The equation in the exercise derived from the gradient is:\[ \frac{\pi \sqrt{3}}{24}(x - 2) - \frac{3\sqrt{3}}{2}(y - \frac{\pi}{12}) - \frac{\pi \sqrt{3}}{24}(z - 4) = 0 \]

This equation helps us visualize the local flat approximation of our surface near point P.
In essence, the tangent plane represents the "best fit" flat surface at that particular point, providing valuable local linearization.

Normal Lines

Normal lines are like skyscrapers rising straight up from our surface at point P. They are perpendicular to the tangent plane and use the gradient vector as their guide.
In three-dimensional space, the normal line is defined using the direction of the gradient because the gradient is always perpendicular to the level surfaces of the function.

• Our problem gives the normal line equation as:
  \[ x = 2 + \frac{\pi \sqrt{3}}{24}t, \, y = \frac{\pi}{12} - \frac{3\sqrt{3}}{2}t, \, z = 4 - \frac{\pi \sqrt{3}}{24}t \]

This shows how any point on the normal line changes as t varies, following the direction dictated by the gradient.
Normal lines are vital in applications ranging from physics to computer graphics, indicating directionality off complex surfaces.