Dylan's questions at Yahoo Answers regarding related rates

[MarkFL]

Here are the questions:

Calculus Review help?

I have a test coming up and these were a few problems I'm having trouble with covering related rates. Help on these problems would be incredibly appreciated.

1. A street light is at the top of a 10.000 ft. tall pole. A man 5.000 ft tall walks away from the pole with a speed of 5.000 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 31.000 feet from the pole?

2.Water is leaking out of an inverted conical tank at a rate of 8800.000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 7.000 meters and the diameter at the top is 4.000 meters. If the water level is rising at a rate of 29.000 centimeters per minute when the height of the water is 1.500 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

3. Gravel is being dumped from a conveyor belt at a rate of 50 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other. How fast is the height of the pile increasing when the pile is 13 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V = [ 1/3]πr2h.

(I got the answer of 0.09417452253 but was incorrect)

4. A plane that is flying horizontally at an altitude of 6 kilometers and a speed of 570 kilometers per hour passes directly over a radar station. How fast is the distance between the plane and the radar station increasing when the distance between the two is 19 kilometers

5. A spherical snowball melts in such a way that its surface area decreases at a rate of 1.4 cm2/min. Find the rate at which its diameter is decreasing when the diameter is 9 cm.

(Note: The surface area of a sphere 4πr2.)
(I got the answer of 0.01733020491 but was incorrect)

6.Two cars start moving from the same point. One travels east at 35 miles per hour and the other travels north at 70 miles per hour. How fast is the distance between them increasing 35 minutes after they start.

7.A trough is 9 feet long and has ends that are isosceles triangles that are 1 foot high and 4 feet wide. If the trough is being filled at a rate of 10 cubic feet per minute, how fast is the height of the water increaseing when the height is 8 inches?

8.According to Boyle's Law, when a sample of gas is compressed at a constant temperature, the pressure and volumn satisfy the equation P V = C, where C is a constant. Assume that, at a certain instant, a sample has a volume of 750 cm3, is a pressure of 160 kPa, and the pressure is increasing at a rate of 18 kPa/min. At what rate is the volume decreasing at this instant?

9.A lighthouse is located on a small island 2 km away from the nearest point P on a straight shoreline. Its light makes 5 revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1.4 kilometers from P?

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Dylan,

1. A street light is at the top of a 10.000 ft. tall pole. A man 5.000 ft tall walks away from the pole with a speed of 5.000 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 31.000 feet from the pole?

Please refer to the following diagram:

View attachment 1581

$P$ is the height of the pole, $M$ is the height of the man, $D$ is the man's distance from the pole, and $x$ is the length of the shadow.

By similarity, we find:

\(\displaystyle \frac{D+x}{P}=\frac{x}{M}\)

\(\displaystyle DM+Mx=Px\)

\(\displaystyle x=\frac{DM}{P-M}\)

Observing that $x$ and $D$ are changing with time, we find by implcitly differentiating with respect to time $t$, we have:

\(\displaystyle \frac{dx}{dt}=\frac{M}{P-M}\frac{dD}{dt}\)

Also observing that movement of the tip of the man's shadow is the result not only of the shadow growing but also of the man's movement, we find that we must add these rates of change, so that the rate of change of the point $T$ of the tip of the shadow is given by:

\(\displaystyle \frac{dT}{dt}=\frac{dD}{dt}+\frac{dx}{dt}=\frac{dD}{dt}\left(1+\frac{M}{P-M} \right)=\frac{P}{P-M}\frac{dD}{dt}\)

Now, using the given data (did you notice we do not need $D$?):

\(\displaystyle P=10\text{ ft},\,M=5\text{ ft},\,\frac{dD}{dt}=5\,\frac{\text{ft}}{\text{s}}\)

We find:

\(\displaystyle \frac{dT}{dt}=\frac{10}{10-5}\cdot5\,\frac{\text{ft}}{\text{s}}=10\, \frac{\text{ft}}{\text{s}}\)

2.Water is leaking out of an inverted conical tank at a rate of 8800.000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 7.000 meters and the diameter at the top is 4.000 meters. If the water level is rising at a rate of 29.000 centimeters per minute when the height of the water is 1.500 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Let all linear measures be in centimeters and time be measured in minutes.

The statement:

"Water is leaking out of an inverted conical tank at a rate of 8800 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate."

may be expressed mathematically as:

\(\displaystyle \frac{dV}{dt}=R-8800\)

Where $V$ is the volume of the water in the tank at time $t$ and $R$ is the constant rate at which water is being pumped into the tank. $R$ is the quantity we are asked to find.

Using the volume of a cone, we know the volume of the water in the tank may be given by:

\(\displaystyle V=\frac{1}{3}\pi r^2h\)

where $r$ is the radius of the surface of the water, and $h$ is the depth of the water. We know that at any given time or volume of water, the ratio of the radius of the water at the surface to its depth will remain constant, and in fact will be in the same proportions as the tank itself.

Because we are given information regarding the time rate of change of the depth and the depth itself, we need to replace the radius with a function of the depth. Hence, by similarity, we may use:

\(\displaystyle \frac{r}{h}=\frac{2}{7}\implies r=\frac{2h}{7}\)

And so the volume as a function of $h$ is:

\(\displaystyle V=\frac{1}{3}\pi \left(\frac{2h}{7} \right)^2h=\frac{4\pi}{147}h^3\)

Now, differentiating with respect to $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=\frac{4\pi}{49}h^2\frac{dh}{dt}\)

Equating the two expressions for \(\displaystyle \frac{dV}{dt}\), we find:

\(\displaystyle \frac{4\pi}{49}h^2\frac{dh}{dt}=R-8800\)

Solving for $R$, we get:

\(\displaystyle R=\frac{4\pi}{49}h^2\frac{dh}{dt}+8800\)

Now, using the given data (making sure all of our units match):

\(\displaystyle h=150\text{ cm},\,\frac{dh}{dt}=29\,\frac{\text{cm}}{\text{min}}\)

We have:

\(\displaystyle R=\frac{4\pi}{49}\left(150\text{ cm} \right)^2\left(29\,\frac{\text{cm}}{\text{min}} \right)+8800\,\frac{\text{cm}^3}{\text{min}}\)

\(\displaystyle R=\frac{400}{49}\left(6525\pi+1078 \right)\,\frac{\text{cm}^3}{\text{min}}\)

3. Gravel is being dumped from a conveyor belt at a rate of 50 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other. How fast is the height of the pile increasing when the pile is 13 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V = [ 1/3]πr2h.

Let all linear measures be in feet, and time in minutes.

The statement:

"Gravel is being dumped from a conveyor belt at a rate of 50 cubic feet per min"

tells us regarding the rate of change of the volume of the pile of gravel:

\(\displaystyle \frac{dV}{dt}=50\frac{\text{ft}^3}{\text{min}}\)

That is, the volume of the pile is increasing at a rate of 50 cubic feet per minute.

The statement:

"It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other"

tells us:

\(\displaystyle V=\frac{1}{3}\pi \left(\frac{h}{2} \right)^2(h)=\frac{1}{12}\pi h^3\)

This comes from the formula for the volume of a cone, where the base radius is equal to half the height.

Now, if we implicitly differentiate this equation with respect to time $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}\)

Since we are asked to find how fast the height of the pile is increasing, we want to solve for \(\displaystyle \frac{dh}{dt}\):

\(\displaystyle \frac{dh}{dt}=\frac{4}{\pi h^2}\frac{dV}{dt}\)

Now, using the given data:

\(\displaystyle \frac{dV}{dt}=50\frac{\text{ft}^3}{\text{min}},\,h=13\text{ ft}\)

we find:

\(\displaystyle \frac{dh}{dt}=\frac{4}{\pi \left(13\text{ ft} \right)^2}\left(50\frac{\text{ft}^3}{\text{min}} \right)=\frac{200}{169\pi}\frac{\text{ft}}{\text{min}}\)

 

[MarkFL]

4. A plane that is flying horizontally at an altitude of 6 kilometers and a speed of 570 kilometers per hour passes directly over a radar station. How fast is the distance between the plane and the radar station increasing when the distance between the two is 19 kilometers?

First, let's draw a diagram:

View attachment 1582

The plane is at $P$, the radar station is at $R$, $h$ is the altitude of the plane (which is constant since its flight is said to be horizontal) and $x$ is the distance from the radar station to the point on the ground (or at the same level as the radar station) directly below the plane. $s$ is the distance from the plane the the radar station.

Using the Pythagorean theorem, we may state:

(1) \(\displaystyle x^2+h^2=s^2\)

Implicitly differentiating (1) with respect to time $t$, we find:

\(\displaystyle 2x\frac{dx}{dt}=2s\frac{ds}{dt}\)

We are interested in solving for \(\displaystyle \frac{ds}{dt}\) since we are asked to find the rate at which the distance from the plane to the station is increasing when it is 19 km away from the station. So, we find:

\(\displaystyle \frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}\)

Now, we do not know $x$ but we know $h$ and $s$, and so solving (1) for $x$ (and taking the positive root since it represents a distance), we find:

\(\displaystyle x=\sqrt{s^2-h^2}\)

And so we have:

\(\displaystyle \frac{ds}{dt}=\frac{\sqrt{s^2-h^2}}{s}\frac{dx}{dt}\)

Now, the speed $v$ of the plane represents the time rate of change of $x$, hence:

\(\displaystyle v=\frac{dx}{dt}\)

and so we may write:

\(\displaystyle \frac{ds}{dt}=\frac{v\sqrt{s^2-h^2}}{s}\)

Now, we may plug in the given data:

\(\displaystyle v=470\frac{\text{mi}}{\text{hr}},\,s=19\text{ km},\,h=6\text{ km}\)

and we then find:

\(\displaystyle \frac{ds}{dt}=\frac{570\sqrt{19^2-6^2}}{19} \frac{\text{km}}{\text{hr}}=150\sqrt{13} \frac{\text{km}}{\text{hr}}\)

5. A spherical snowball melts in such a way that its surface area decreases at a rate of 1.4 cm^2/min. Find the rate at which its diameter is decreasing when the diameter is 9 cm.

The statement:

"A spherical snowball melts in such a way that its surface area decreases at a rate of 1.4 cm^2/min"

allows us to write:

\(\displaystyle \frac{dS}{dt}=-\frac{7}{5}\,\frac{\text{cm}^2}{\text{min}}\)

where $S$ is the surface area of the spherical snowball at time $t$.

The surface area of a sphere is given by:

\(\displaystyle S=4\pi r^2=\pi(2r)^2=\pi D^2\)

where $D=2r$ is the diameter of the sphere.

Differentiating with respect to time $t$, we find:

\(\displaystyle \frac{dS}{dt}=2\pi D\frac{dD}{dt}\)

We are asked to find the time rate of change of the diameter, so we want to solve for \(\displaystyle \frac{dD}{dt}\):

\(\displaystyle \frac{dD}{dt}=\frac{1}{2\pi D}\frac{dS}{dt}\)

Now, using the given data:

\(\displaystyle D=9\text{ cm},\,\frac{dS}{dt}=-\frac{7}{5}\,\frac{\text{cm}^2}{\text{min}}\)

we find:

\(\displaystyle \left. \frac{dD}{dt} \right|_{D=9\text{ cm}}=\frac{1}{2\pi\left(9\text{ cm} \right)}\left(-\frac{7}{5}\,\frac{\text{cm}^2}{\text{min}} \right)=-\frac{7}{90\pi}\,\frac{\text{cm}}{\text{min}}\)

6.Two cars start moving from the same point. One travels east at 35 miles per hour and the other travels north at 70 miles per hour. How fast is the distance between them increasing 35 minutes after they start.

Let's orient our coordinate axes such that the starting point for the cars is the origin. At time $t$ in hours, the position of the car traveling east is $\left(v_1t,0 \right)$ and the position of the car traveling north is $\left(0,v_2t \right)$

Thus, using the distance formula, we find the distance $D$ between them is:

\(\displaystyle D(t)=t\sqrt{v_1^2+v_2^2}\)

Differentiating with respect to $t$, we find:

\(\displaystyle \frac{dD}{dt}=\sqrt{v_1^2+v_2^2}\)

We find then that the time rate of change of the distance between the cars is a constant, depending only on the speeds of the two cars. If we write:

\(\displaystyle v_2=kv_1\)

Then we have:

\(\displaystyle \frac{dD}{dt}=v_1\sqrt{1+k^2}\)

In our case, we have:

\(\displaystyle v_1=35\,\frac{\text{mi}}{\text{hr}},\,k=2\)

Hence:

\(\displaystyle \frac{dD}{dt}=35\sqrt{5}\,\frac{\text{mi}}{\text{hr}}\)

 

[MarkFL]

7.A trough is 9 feet long and has ends that are isosceles triangles that are 1 foot high and 4 feet wide. If the trough is being filled at a rate of 10 cubic feet per minute, how fast is the height of the water increasing when the height is 8 inches?

We need to find the volume $V$ of water in the trough as a function of its depth $h$. Let $w$ be the width of the triangular cross-sections of the water in the trough. The area $A$ of a cross-section can be found using the formula for the area of a triangle:

\(\displaystyle A=\frac{1}{2}wh\)

We know the width $w$ increases linearly as $h$ increases, and is $0$ when $h=0$ and is $4$ when $h=1$, hence:

\(\displaystyle w=4h\)

So we find:

\(\displaystyle A=\frac{1}{2}(4h)h=2h^2\)

Letting $L$ be the length of the trough, the volume of water may be given as:

\(\displaystyle V=AL=2Lh^2\)

Differentiating with respect to time $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=4Lh\frac{dh}{dt}\)

Since we are being asked about the rate of change of the depth of the water, we want to solve for \(\displaystyle \frac{dh}{dt}\):

\(\displaystyle \frac{dh}{dt}=\frac{1}{4Lh}\frac{dV}{dt}\)

Now, we are given:

\(\displaystyle L=9\text{ ft},\,h=\frac{2}{3}\,\text{ft},\,\frac{dV}{dt}=10\,\frac{\text{ft}^3}{\text{min}}\)

And so we find:

\(\displaystyle \frac{dh}{dt}=\frac{1}{4\left(9\text{ ft} \right)\left(\frac{2}{3}\,\text{ft} \right)}\left(10\,\frac{\text{ft}^3}{\text{min}} \right)=\frac{5}{12}\,\frac{\text{ft}}{\text{min}}\)

8.According to Boyle's Law, when a sample of gas is compressed at a constant temperature, the pressure and volume satisfy the equation PV = C, where C is a constant. Assume that, at a certain instant, a sample has a volume of 750 cm^3, is at a pressure of 160 kPa, and the pressure is increasing at a rate of 18 kPa/min. At what rate is the volume decreasing at this instant?

Let's begin with the equation given by Boyle's Law:

\(\displaystyle PV=C\)

Implcitly differentiating with respect to time $t$, we find:

\(\displaystyle P\frac{dV}{dt}+\frac{dP}{dt}V=0\)

Since we are asked to find the time rate of change of the volume, we want to solve for \(\displaystyle \frac{dV}{dt}\):

\(\displaystyle \frac{dV}{dt}=-\frac{dP}{dt}\frac{V}{P}\)

Using the given data:

\(\displaystyle \frac{dP}{dt}=18\frac{\text{kPa}}{\text{min}},\,V=750\text{ cm}^3,\,P=160\text{ kPa}\)

we find:

\(\displaystyle \frac{dV}{dt}=-\left(18\frac{\text{kPa}}{\text{min}} \right)\frac{750\text{ cm}^3}{160\text{ kPa}}=-\frac{675}{8}\,\frac{\text{cm}^3}{\text{min}}\)

9.A lighthouse is located on a small island 2 km away from the nearest point P on a straight shoreline. Its light makes 5 revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1.4 kilometers from P?

Let's generalize a bit and derive a formula we can then plug our data into.

The first thing I would do is draw a diagram:

View attachment 1583

As we can see, we may state:

$\displaystyle \tan(\theta)=\frac{x}{y}$

Now, let's differentiate with respect to time $t$, bearing in mind that while $x$ is a function of $t$, $y$ is a constant.

$\displaystyle \sec^2(\theta)\cdot\frac{d\theta}{dt}=\frac{1}{y} \cdot\frac{dx}{dt}$

Since we are being asked to find the speed of the spot, whose position is $x$, we want to solve for $\dfrac{dx}{dt}$:

$\displaystyle \frac{dx}{dt}=y\sec^2(\theta) \cdot\frac{d\theta}{dt}$

Let's let the angular velocity be given by:

$\omega=\dfrac{d\theta}{dt}$

and from the diagram and the Pythagorean theorem, we find:

$\displaystyle \sec^2(\theta)=\frac{x^2+y^2}{y^2}$

Hence, we have:

$\displaystyle \frac{dx}{dt}=\frac{\omega}{y}\left(y^2+x^2 \right)$

Now we may plug in the given data:

$\displaystyle \omega=10\pi\frac{1}{\text{min}},\,y=2\text{ km},\,x=\frac{7}{5}\text{ km}$

And so we find:

$\displaystyle \frac{dx}{dt}=\frac{10\pi}{2}\left(2^2+\left(\frac{7}{5} \right)^2 \right)\, \frac{\text{km}}{\text{min}}=\frac{149 \pi}{5}\, \frac{\text{km}}{\text{min}}$

 

[MarkFL]

Dylan's response:

Incredibly helpful, thank you very much sir. May I ask how you got those answers so quickly?

The comment function there is not working for me, so I am responding here:

Hello Dylan,

For some of these problems, I was able to take solutions to similar problems I have posted here and modify them to fit your questions. For the rest, I just worked the problems. I have been doing related rates problems for many years, and with experience it just gets easier.

Best Regards,

Mark.