Related Rates

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April 10th 2009, 09:07 AM
Juggalomike

Related Rates

Ok i have 2 problems that i am stuck on. Please show me how to work this out as compared to simply posting the answer, thanks.

1. A light is affixed to the top of a 12 foot tall lamp-post. A 6 foot tall man walks away from the lamp post at a rate of 5 ft/sec. How fast i the length of his shadow increasing when he is 5 feet away?

Work so far: I set ds/dt as 5 ft/s because it is the rate of change. I am thinking that to solve this i should break it down into two triangles but i am lost as to how i can do that because i have no idea how to get the current distance of his shadow.

2. A water tank has the shape of an inverted right circular cone of altitude 12 feet and base radius 6 feet. If water is being pumped into the tank at a rate of 10 gal/min, at what rate is the water level rising when the water is 3 feet deep?

Work so far: Again i set dh/dt at 10 gal/min. I believe i should use (πr^2h)/3 but i have no idea what to sub in for R or H and how to solve it.

Update: For number 2 i believe dv/dt is 45π/2, i came to this conclusion by solving for V=(1/3)π(6h/12)^2(h), however i am not sure if this is correct

Any help anyone can provide is greatly appreciated, thanks
April 10th 2009, 09:53 AM
stapel

Quote:

Originally Posted by Juggalomike

1. A light is affixed to the top of a 12 foot tall lamp-post. A 6 foot tall man walks away from the lamp post at a rate of 5 ft/sec. How fast i the length of his shadow increasing when he is 5 feet away?

Draw the post and the man. Draw the ground. Draw the slanty line from the top of the post, across the top of the man's head, and down to the ground. This gives you two nested (and thus similar) right triangles.

Label the post as "12", the man as "6", and the ground between the post and the man as "x", and note that dx/dt = 5.

Label the ground between the man and the tip of the triangle as "y". You need to find dy/dt.

The similar triangles give you the proportion 6/y = 12/(x + y). Use this to find the value of y when x = 5.

Differentiate the proportion with respect to time "t". Plug in the known values, and solve for the value of dy/dt.

Quote:

Originally Posted by Juggalomike

2. A water tank has the shape of an inverted right circular cone of altitude 12 feet and base radius 6 feet. If water is being pumped into the tank at a rate of 10 gal/min, at what rate is the water level rising when the water is 3 feet deep?

Draw the upside-down cone. Draw a horizontal line indicating the water level. Draw the vertical "height" line. Note that, regarding one side or half of the cone's cross-sectional view, you now have two nested (and thus similar) right triangles.

Label the total height as "12" and the radius across the top as "6"; note that dV/dt = 10. Label the water height as "h" and the radius of the water's height as "r".

The similar triangles give you the proportion 12/6 = h/r. Use this to find the value of "r" when h = 3.

Solve the proportion for r in terms of h. Plug the result into the formula the formula for the volume of the cone. Differentiate with respect to time "t". Plug in all the known values, and solve for dh/dt.

If you get stuck, please reply showing how far you have gotten. Thank you! (Wink)
April 10th 2009, 03:35 PM
Soroban

Hello, Juggalomike!

Quote:

1. A light is affixed to the top of a 12 foot tall lamppost.
A 6-foot tall man walks away from the lamp post at a rate of 5 ft/sec.
How fast is the length of his shadow increasing when he is 5 feet away?

Code:

A * | * C | * 12 | | * | 6 | * | | * * - - - * - - - - - - - * - - - B x D s E

The lampost is: $AB = 12$
The man is: $CD = 6$
His distance from the lampost is: $x = BD$ , and $\frac{dx}{dt} = 5$
The length of his shadow is: $s = DE$

Since $\Delta CDE \sim \Delta ABE\!:\;\;\frac{s}{6} \:=\:\frac{x+s}{12} \quad\Rightarrow\quad s \:=\:x$

Differentiate with respect to time: . $\frac{ds}{dt} \:=\:\frac{dx}{dt}$

Therefore: . $\frac{ds}{dt} \:=\:5$

His shadow is lengthening at 5 ft/min.

Quote:

2. A water tank has the shape of an inverted right circular cone
of altitude 12 feet and base radius 6 feet.
If water is being pumped into the tank at a rate of 10 gal/min,
at what rate is the water level rising when the water is 3 feet deep?

Nasty! .Nasty! .They give the volume in gallons, then ask for answers in feet.
. . They could have given us a conversion formula at the very least!
I had to look it up: . $1\text{ gal} \:\approx\:0.1337\text{ ft}^3$

So we have: . $\frac{dV}{dt} \:=\:10\text{ gal/min} \:=\:1.337\text{ ft}^3\text{/min}$

Look at the side view of the cone:

Code:

A 6 D - * - - - - * - - - - * : \ | / : \ | / : \ B| r / : * - - + - - * E 12 \ | / : \ | / : \ h| / : \ | / : \|/ - * C

The height of the cone is: $AC = 12$
The radius of the cone is: $AD = 6$

The height of the water is: $h = BC$
The radius of the water is: $r = BE$

Since $\Delta EBC \sim \Delta DAC\!:\;\;\frac{r}{h} \:=\:\frac{6}{12}\quad\Rightarrow\quad r \:=\:\frac{h}{2}\;\;{\color{blue}[1]}$

The volume of the water is: . $V \:=\:\frac{\pi}{3}r^2h$

Substitute [1]: . $V \:=\:\frac{\pi}{3}\left(\frac{h}{2}\right)^2h \quad\Rightarrow\quad V \:=\:\frac{\pi}{12}h^3$

Differentiate with respect to time: . $\frac{dV}{dt}\:=\:\frac{\pi}{4}h^2\,\frac{dh}{dt}$

Substitute our known values: . $1.337 \:=\:\frac{\pi}{4}(3^2)\,\frac{dh}{dt}$

Therefore: . $\frac{dh}{dt} \;=\;\frac{4}{9\pi}(1.337) \;\approx\;0.05\text{ ft/min}$
April 10th 2009, 04:12 PM
Juggalomike

Quote:

Originally Posted by stapel

Draw the post and the man. Draw the ground. Draw the slanty line from the top of the post, across the top of the man's head, and down to the ground. This gives you two nested (and thus similar) right triangles.

Label the post as "12", the man as "6", and the ground between the post and the man as "x", and note that dx/dt = 5.

Label the ground between the man and the tip of the triangle as "y". You need to find dy/dt.

The similar triangles give you the proportion 6/y = 12/(x + y). Use this to find the value of y when x = 5.

Differentiate the proportion with respect to time "t". Plug in the known values, and solve for the value of dy/dt.

Draw the upside-down cone. Draw a horizontal line indicating the water level. Draw the vertical "height" line. Note that, regarding one side or half of the cone's cross-sectional view, you now have two nested (and thus similar) right triangles.

Label the total height as "12" and the radius across the top as "6"; note that dV/dt = 10. Label the water height as "h" and the radius of the water's height as "r".

The similar triangles give you the proportion 12/6 = h/r. Use this to find the value of "r" when h = 3.

Solve the proportion for r in terms of h. Plug the result into the formula the formula for the volume of the cone. Differentiate with respect to time "t". Plug in all the known values, and solve for dh/dt.

If you get stuck, please reply showing how far you have gotten. Thank you!

Ok i got to the last part in problem 1 where you say "Differentiate the proportion with respect to time "t". Plug in the known value, and solve for value of dy/dt. What equation do i differentiate? I solved 6/y=12/(x+y) and figured out that y=5.

I am also stuck on problem 2 where you say "solve the proportion for r in terms of h". I am unfortunately terrible with terminology, what exactly does this mean? thanks
April 10th 2009, 05:32 PM
stapel

Quote:

Originally Posted by Juggalomike

Ok i got to the last part in problem 1 where you say "Differentiate the proportion with respect to time "t". Plug in the known value, and solve for value of dy/dt. What equation do i differentiate?

"The proportion". (Wink)

Quote:

Originally Posted by Juggalomike

I am also stuck on problem 2 where you say "solve the proportion for r in terms of h". I am unfortunately terrible with terminology, what exactly does this mean?

They were supposed to have taught you this back in algebra! (Surprised)

To learn what proportions are and how to solve them, try here.
April 10th 2009, 08:37 PM
Juggalomike

Quote:

Originally Posted by stapel

"The proportion". (Wink)

They were supposed to have taught you this back in algebra! (Surprised)

To learn what proportions are and how to solve them, try here.

haha after reading that i did learn it and still know how to solve it, the term was what was throwing me off, took algebra 2 almost 6 years ago, forgot what the word itself meant, but thanks to both of you for your help
April 12th 2009, 06:35 PM
Juggalomike

Ok 2 questions, the first is in regards to the 2nd problem.

Question 1.

This is my work, i am getting a different answer then you for some reason

DV/DT=π/3[r^2*DH/DT+2rh(1/2*DH/DT)]
DV/DT=π/3[r^2*DH/DT+rh*DH/DT]
DV/DT=π/3*DH/DT*(r^2+rh)
10=π/3*DH/DT*[(3/2)^2+(3/2*3)]
10=π/3*DH/DT*(9/4+18/4)
10=9π/4*DH/DT

40/9π=40/9*3.14=40/28.26=1.4

Question 2

I got the answer i would just like someone to check it if at all possible.

Q. If the area of a circle is increasing at a constant rate of 200π square inches/sec, find the rate of increase of the circumference when the circumference is 100π inches.

My Work:

DA/DT= 200π in^/sec
A=πr^2
C=2πr
DC/DT=?

2A=Cr
2DA/DT=r*DC/DT+C*DR/DT
2DA/DT=r*DC/DT+c/2π*DC/DT
2DA/DT=DC/DT*[r+c/2π]
2*200π=DC/DT*[50+100π/2π]
400π=DC/DT*100
DC/DT=4π in/sec