Related Rates: Learn It 2

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Related Rates: Learn It 2

Related-Rates Problem-Solving

Examples of the Process

Let’s now implement the strategy just described to solve several related-rates problems. The first example involves a plane flying overhead. The relationship we are studying is between the speed of the plane and the rate at which the distance between the plane and a person on the ground is changing.

An airplane is flying overhead at a constant elevation of $4000$ ft. A man is viewing the plane from a position $3000$ ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of $600$ ft/sec, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

Step 1. Draw a picture, introducing variables to represent the different quantities involved.

A right triangle is made with a person on the ground, an airplane in the air, and a radio tower at the right angle on the ground. The hypotenuse is s, the distance on the ground between the person and the radio tower is x, and the side opposite the person (that is, the height from the ground to the airplane) is 4000 ft. — Figure 2. An airplane is flying at a constant height of 4000 ft. The distance between the person and the airplane and the person and the place on the ground directly below the airplane are changing. We denote those quantities with the variables $s$ and $x$ , respectively.

As shown, $x$ denotes the distance between the man and the position on the ground directly below the airplane. The variable $s$ denotes the distance between the man and the plane.

Note that both $x$ and $s$ are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of $4000$ ft. Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length $4000$ ft is perpendicular to the line segment of length $x$ feet, creating a right triangle.

Step 2. Since $x$ denotes the horizontal distance between the man and the point on the ground below the plane, $dx/dt$ represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, $\frac{dx}{dt}=600$ ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find $ds/dt$ when $x=3000$ ft.

Step 3. From Figure 2, we can use the to write an equation relating $x$ and $s$ :

$[x(t)]^2+4000^2=[s(t)]^2$ .

Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

$x\frac{dx}{dt}=s\frac{ds}{dt}$ .

Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find $\frac{ds}{dt}$ when $x=3000$ ft.

Since the speed of the plane is $600$ ft/sec, we know that $\frac{dx}{dt}=600$ ft/sec. We are not given an explicit value for $s$ ; however, since we are trying to find $\frac{ds}{dt}$ when $x=3000$ ft, we can use the Pythagorean theorem to determine the distance $s$ when $x=3000$ and the height is $4000$ ft. Solving the equation

3000^{2} + 4000^{2} = s^{2}

$3000^2+4000^2=s^2$

for $s$ , we have $s=5000$ ft at the time of interest. Using these values, we conclude that $ds/dt$ is a solution of the equation

(3000) (600) = (5000) \cdot \frac{d s}{d t}

$(3000)(600)=(5000) \cdot \frac{ds}{dt}$ .

Therefore,

$\frac{ds}{dt}=\frac{3000 \cdot 600}{5000}=360$ ft/sec.

Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities $x(t)$ and $s(t)$ by the equation

[x (t)]^{2} + 4000^{2} = [s (t)]^{2}

$[x(t)]^2+4000^2=[s(t)]^2$ .

Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted $x(t)=3000$ into the equation before differentiating, our equation would have been

3000^{2} + 4000^{2} = [s (t)]^{2}

$3000^2+4000^2=[s(t)]^2$ .

After differentiating, our equation would become

0 = s (t) \frac{d s}{d t}

$0=s(t)\frac{ds}{dt}$ .

As a result, we would incorrectly conclude that $\frac{ds}{dt}=0$ .

Watch the following video to see the worked solution to this example/

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “4.1 Related Rates” here (opens in new window).

What is the speed of the plane if the distance between the person and the plane is increasing at the rate of $300$ ft/sec?

$\frac{ds}{dt}=300$ ft/sec

$500$ ft/sec

What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of $4000$ ft from the launch pad and the velocity of the rocket is $500$ ft/sec when the rocket is $2000$ ft off the ground?

Find $\frac{d\theta}{dt}$ when $h=2000$ ft. At that time, $\frac{dh}{dt}=500$ ft/sec.

$\frac{1}{10}$ rad/sec

In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.

Water is draining from the bottom of a cone-shaped funnel at the rate of $0.03 \, \text{ft}^3 /\text{sec}$ . The height of the funnel is $2$ ft and the radius at the top of the funnel is $1$ ft. At what rate is the height of the water in the funnel changing when the height of the water is $\frac{1}{2}$ ft?

Step 1: Draw a picture introducing the variables.

A funnel is shown with height 2 and radius 1 at its top. The funnel has water to height h, at which point the radius is r. — Figure 3. Water is draining from a funnel of height 2 ft and radius 1 ft. The height of the water and the radius of water are changing over time. We denote these quantities with the variables $h$ and $r,$ respectively.

Let $h$ denote the height of the water in the funnel, $r$ denote the radius of the water at its surface, and $V$ denote the volume of the water.

Step 2: We need to determine $\frac{dh}{dt}$ when $h=\frac{1}{2}$ ft. We know that $\frac{dV}{dt}=-0.03 \text{ft}^3 / \text{sec}$ .

Step 3: The volume of water in the cone is

V = \frac{1}{3} π r^{2} h

$V=\frac{1}{3}\pi r^2 h$

From Figure 3, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, $\frac{r}{h}=\frac{1}{2}$ or $r=\frac{h}{2}$ . Using this fact, the equation for volume can be simplified to

$V=\frac{1}{3}\pi (\frac{h}{2})^2 h=\frac{\pi}{12}h^3$

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time $t$ , we obtain

$\frac{dV}{dt}=\frac{\pi}{4}h^2 \frac{dh}{dt}$

Step 5: We want to find $\frac{dh}{dt}$ when $h=\frac{1}{2}$ ft. Since water is leaving at the rate of $0.03 \, \text{ft}^3 / \text{sec}$ , we know that $\frac{dV}{dt}=-0.03 \, \text{ft}^3 / \text{sec}$ . Therefore,

- 0.03 = \frac{π}{4} (\frac{1}{2})^{2} \frac{d h}{d t}

$-0.03=\frac{\pi}{4}(\frac{1}{2})^2 \frac{dh}{dt}$ ,

which implies

- 0.03 = \frac{π}{16} \frac{d h}{d t}

$-0.03=\frac{\pi}{16}\frac{dh}{dt}$

It follows that

\frac{d h}{d t} = - \frac{0.48}{π} = - 0.153

$\frac{dh}{dt}=-\frac{0.48}{\pi}=-0.153$ ft/sec.

At what rate is the height of the water changing when the height of the water is $\frac{1}{4}$ ft?

We need to find

\frac{d h}{d t}

$\frac{dh}{dt}$ when

h = \frac{1}{4}

$h=\frac{1}{4}$ .

$-0.61$ ft/sec