PAGE 1

3.11 Related Rates (part 2)

Example 1. Two airplanes take off from an airport. One leaves at noon flying 500 mi/hr due east. The other one leaves at 1 pm flying 600 mi/hr due north. Assuming both planes fly at the same and constant altitude, how fast is the distance between them changing at 3 pm?

Figure: Coordinate axes with 'airport' at origin, 'airplane flying north' on y-axis, and 'airplane flying east' on x-axis.

Figure: Right triangle diagram with vertical leg y, horizontal leg x, and hypotenuse z.

x: dist of the plane flying east from airport
y: dist of the plane flying north from airport
z: dist between them

PAGE 2

Known: \( \frac{dx}{dt} = 500 \) , \( \frac{dy}{dt} = 600 \)

want: \( \frac{dz}{dt} \) at 3 pm

\( z^2 = x^2 + y^2 \) \( x, y, z \) are functions of \( t \)

\( \frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2) \)

\( 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \)

\( \frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z} \)

\( = \frac{1500(500) + 1200(600)}{\sqrt{1500^2 + 1200^2}} \)

\( = \) 765.25 miles/hour

at 3pm, \( \frac{dx}{dt} = 500 \), \( \frac{dy}{dt} = 600 \)

\( x = 500 \cdot 3 = 1500 \) (left at noon, flying at 500 for 3 hrs)

\( y = 600 \cdot 2 = 1200 \)

\( z = \sqrt{x^2 + y^2} = \sqrt{1500^2 + 1200^2} \)

PAGE 3

Related Rates: Airplane Example

Example 2. An airplane is flying horizontally at an altitude of 1 mile and a speed of 500 mph. An observer on the ground is looking at the plane. At what rate is the angle between the line from the observer to the airplane and the ground changing when the airplane is 2 miles from the observer?

Figure: A right triangle diagram with labels 'observer', 'ground', and 'airplane' with an arrow indicating horizontal flight.

Variables:

z: dist. from observer to plane
x: dist. from observer to point below airplane on the ground
y: altitude (\( = 1 \))
θ: angle of elevation

PAGE 4

Known: \( y = 1 \)

\( \frac{dx}{dt} = 500 \)

Want: \( \frac{d\theta}{dt} \) when \( z = 2 \)

We want to relate \( \theta, x, z \) (adjacent and hypotenuse)

\[ \cos \theta = \frac{\text{adj}}{\text{hyp}} = \frac{x}{z} \]

Differentiate with respect to \( t \)

\[ -\sin \theta \cdot \left( \frac{d\theta}{dt} \right) = \frac{z \cdot \frac{dx}{dt} - x \cdot \left( \frac{dz}{dt} \right)}{z^2} \]

(quotient rule used)

We don't know \( \frac{dz}{dt} \) so relating \( x \) and \( z \) is not useful.

Cosine is not good.

Try a different one without hypotenuse.

Let's try tangent.

PAGE 5

Starting with the trigonometric relationship from the diagram:

\[ \tan \theta = \frac{1}{x} = x^{-1} \]

Differentiate with respect to time \( t \):

\[ \sec^2 \theta \cdot \frac{d\theta}{dt} = -x^{-2} \cdot \frac{dx}{dt} \]

Figure: Right triangle with angle theta, opposite side 1, adjacent side x, and hypotenuse z.

Solving for \( \frac{d\theta}{dt} \):

\[ \frac{d\theta}{dt} = \frac{-x^{-2} \frac{dx}{dt}}{\sec^2 \theta} = \frac{-\frac{dx}{dt}}{x^2 \frac{1}{\cos^2 \theta}} = -\frac{\cos^2 \theta \cdot \frac{dx}{dt}}{x^2} \]

When \( z = 2 \)

\[ z^2 = 1 + x^2 \]

\[ x = \sqrt{z^2 - 1} = \sqrt{3} \]

Figure: Right triangle with hypotenuse 2, opposite side 1, adjacent side square root of 3, and angle theta.

Need \( \cos \theta \):

\[ \cos \theta = \frac{\sqrt{3}}{2} \]

Substituting the values back into the derivative equation:

\[ \frac{d\theta}{dt} = -\frac{(\frac{\sqrt{3}}{2})^2 \cdot 500}{(\sqrt{3})^2} = \text{notebox: } -125 \text{ radians/hr} \]

-125

radians/hr

PAGE 6

Example 4. The bottom of a large theater screen is 3 ft above your eye level and the top of the screen is 10 ft above your eye level. Assume you walk toward the screen (perpendicular to the screen) at a rate of 7 ft/s while looking at the screen. What is the rate of change of the viewing angle (from the bottom of the screen to the top of the screen) when you are 70 ft from the wall, assuming the floor is flat?

Figure: Diagram showing a person looking at a theater screen. Heights of 3ft and 10ft are marked, with viewing angle theta.

Variables

\( \theta \): viewing angle changing as you walk toward/away from screen

PAGE 7

Related Rates: Viewing Angle Problem

We define the variables for the viewing angle problem as follows:

̑: angle from ground to bottom of screen
̒: angle from ground to top of screen
x: dist. from you to wall w/ screen

Known:

\[ \frac{dx}{dt} = -7 \]

(↑ negative because walking toward)

Want: \( \frac{d\theta}{dt} \) when \( x = 70 \)

Figure: A right triangle diagram with a vertical side of height 10, a horizontal base x, and angles alpha, beta, and theta.

Setting up the Equation

Notice that the viewing angle \( \theta \) is the difference between the two larger angles:

\[ \theta = \beta - \alpha \]

Using trigonometry for the right triangles:

\[ \tan \beta = \frac{10}{x} \quad \tan \alpha = \frac{3}{x} \]

Solving for the angles:

\[ \beta = \tan^{-1}\left(\frac{10}{x}\right) \quad \alpha = \tan^{-1}\left(\frac{3}{x}\right) \]

Thus, the equation for \( \theta \) is:

\[ \theta = \tan^{-1}\left(\frac{10}{x}\right) - \tan^{-1}\left(\frac{3}{x}\right) \]

Differentiation

Recall the derivative rule:

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \frac{du}{dx} \]

Take the derivative with respect to time \( t \):

\[ \frac{d\theta}{dt} = \frac{1}{1+\left(\frac{10}{x}\right)^2} \frac{d}{dt}\left(\frac{10}{x}\right) - \frac{1}{1+\left(\frac{3}{x}\right)^2} \frac{d}{dt}\left(\frac{3}{x}\right) \]

Note: \( \frac{10}{x} \) can be written as \( 10x^{-1} \) and \( \frac{3}{x} \) as \( 3x^{-1} \) for easier differentiation.

PAGE 8

Final Calculation

Applying the power rule to the inner terms:

\[ \frac{d\theta}{dt} = \frac{1}{1+\left(\frac{10}{x}\right)^2} \left( -10x^{-2} \cdot \frac{dx}{dt} \right) - \frac{1}{1+\left(\frac{3}{x}\right)^2} \left( -3x^{-2} \cdot \frac{dx}{dt} \right) \]

Substitute the known values: \( x = 70 \) and \( \frac{dx}{dt} = -7 \):

\[ \text{sub in } x=70, \frac{dx}{dt}=-7 \]

After simplifying the expression:

\[ \frac{d\theta}{dt} = \dots = \boxed{0.0097} \text{ radians/s} \]

PAGE 9

Example 3: Related Rates Ladder Problem

Example 3: The top of a ladder leaning against a wall is sliding down at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

Figure: Sketch of a ladder leaning against a wall and floor at two positions: 'initially' and 'some time later'.

Figure: Right triangle diagram with sides labeled x, y, and hypotenuse z.

z: length of ladder
x: dist. from bottom of ladder to wall
y: dist. of top of ladder to ground

x, y are functions of time t
z cannot change

Given Information

Known: \( \frac{dy}{dt} = -0.15 \), \( \frac{dx}{dt} = 0.2 \) when \( x = 3 \)

Want: \( z \)

PAGE 10

Solution

\[ x^2 + y^2 = z^2 \]

Differentiate with respect to \( t \):

\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

When \( x = 3 \):

\[ (2)(3)(0.2) + (2)(y)(-0.15) = 0 \]

Need y:

Figure: Right triangle with base 3 and height y.

\( y = 4 \)

Now we know y:

Figure: Right triangle with base 3 and height 4.

\[ z = \sqrt{3^2 + 4^2} = \boxed{5} \]

Ladder is 5 m long