Shadow Lightpost Problem

Category: Differential Calculus, Analytic Geometry

"Published in Suisun City, California, USA"

The figure shows a lamp located three units to the right of the y-axis and a shadow created by the elliptical region x² + 4y² ≤ 5. If point (- 5, 0) is the edge of the shadow, how far above the x-axis is the lamp located?

Photo by Math Principles in Everyday Life

Solution:

As you can see from the figure that the two lines which are the edges of the shadow are tangent to the elliptical region. In this problem, we will use only one line which contains a point (- 5, 0). Let's consider the given equation of the elliptical region

Take the derivative on both sides of the equation with respect to y to get the slope of a curve by implicit differentiation

You cannot substitute (-5, 0) to the above equation because the point (-5, 0) is not in the curve. Let (x, y) be the point of intersection of a line and an ellipse. In short, its tangent point. Get the equation of a tangent line by point-slope form, we have

Substitute (-5, 0) to the above equation

Equate the slope of a line with the slope of a curve

Substitute the above equation to the equation of an ellipse

Solve for the value of y

Since their point of intersection or tangent point is located in second quadrant, choose the positive value which is y = 1. The tangent point is (- 1, 1). The slope of a curve at (- 1, 1) is

The slope of a curve is the same as the slope of a line because the line is tangent to the curve. We can get the equation of a tangent line using the point-slope form

Substitute (- 5, 0) and m = ¼ to the above equation

From the word problem, the lamp is located 3 units from the y-axis which is x = 3. We can solve the height of the lamp from the ground or x-axis by substitute x = 3 to the above equation

Therefore, the height of the lamp from the ground or x-axis is 2 units.  

Rectangular Parallelepiped Problem

Category: Solid Geometry

"Published in Suisun City, California, USA"

A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side x at each corner and then folding up the sides. Express the volume V of the box as a function of x. 

Solution:

To visualize the problem, let's draw the figure first as follows:

Photo by Math Principles in Everyday Life

If you cut and remove the four squares of side x and then fold up the four sides, the figure is a rectangular parallelepiped. 

Photo by Math Principles in Everyday Life

The volume of a rectangular parallelepipied is given by the formula

                             V = L W H

Substitute the given dimensions, we have

                             V = (20 - 2x)(12 - 2x)(x)

                             V = [2(10 - x)][2(6 - x)](x)

                             V = 4x (10 - x)(6 - x)

Therefore,

                         V(x) = 4x (10 - x)(6 - x)

                         where  0 < x < 6 
      

Maximum Volume - Right Circular Cylinder

Category: Differential Calculus, Solid Geometry

"Published in Suisun City, California, USA"

A right circular cylinder is inscribed in a sphere with radius R. Find the largest possible volume of such a cylinder. 

Solution:

To visualize the problem, let's draw the figure first. Inscribed means inside and so a right circular cylinder is located inside the sphere. 

Photo by Math Principles in Everyday Life

Next, we have to find the dimensions of a right circular cylinder in order to get its volume. By labeling the figure further, we have

Photo by Math Principles in Everyday Life

By Pythagorean Theorem,

We know that the volume of a right circular cylinder is

but h = 2a

Equate r² on both sides of the equation,

Take the derivative of both sides of the equation with respect to a and equate it to zero because we want to maximize the volume of a right circular cylinder. Consider R as a constant in the equation.

Now, we can solve for r,

Therefore,