# 4 Calculus Related-Rates Problems

1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dV/dt, the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/dt, the change in the radius with respect to time. Remember that the volume of a sphere is V=4/3 pi r^3.

2. A baseball diamond is a square 90 feet on a side. A player runs from first base to second base at a rate of 15 feet per second. At what rate is the player's distance from third base decreasing when the player is halfway between first and second base? We've already set part of this problem up. If we let x be the distance between the player and second base,and y be the distance between the player and third base, then dx/dt=-15 feet per second, and dy/dt will tell us what we want to know. Use the picture to find a relationship that will help you answer the question.

3. A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per second, and we're looking for dv/dt . Use the picture to help you find the relationship between x and y, and use it to answer the question asked here.

4. Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = pi(r^2)h, where r is the radius and h is the height of this cylinder.)

1. 👍
2. 👎
3. 👁
1. These are all just applications of the chain rule. If y is a function of u, and u is a function of t, then
dy/dt = dy/du * du/dt

So, now we see what do do:
#1
v=4/3 π r^3
dv/dt = 4πr^2 dr/dt
dr/dt = 15/(4πr^2)
As expected, if the volume changes at a steady rate, the bigger the balloon gets, the slower the radius increases.

#2
Using the Pythagorean Theorem, we know that

x^2 + 90^2 = y^2
x dx/dt = y dy/dt
So, find y when x=45, and then just plug in dx/dt = -15

#3
You don't say what y is, but suppose y is the distance of the tip of the shadow from the man. In that case, using similar triangles,

y/2 = (x+y)/5
y = 2/3 x
So,
dy/dt = 2/3 dx/dt

#4
1st way: use the product rule
v = πr^2h
dv/dt = 2πrh dr/dt + πr^2 dh/dt
You know that dv/dt=0, since the volumem of oil is constant.
Now, what is h? πr^2h = 1, so h = 1/(π*.01^2)
Now you can plug in the numbers.

2nd way: use the fact that v is constant
πr^2h = 1, so
πh = 1/r^2
π dh/dt = -2/r^3 dr/dt
now just plug in your numbers to find dh/dt

don't forget to watch the units.
1m^3 = 10^6 cm^3

1. 👍
2. 👎
2. Steve is tha goat

1. 👍
2. 👎
3. Thanks bestie Steve! Hope ur thriving bc im not! <3

1. 👍
2. 👎

## Similar Questions

1. ### Calculus

a spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. how fast is the radius of the balloon increasing at the instant the radius is 30 centimeters?

2. ### Calculus

Air is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm? V= 4/3*pi*r^3 S= 4 pi r^2

3. ### calculus

A spherical balloon is losing air at the rate of 2 cubic inches per minute. How fast is the radius of the ballon shrinking when the radius is 8 inches.

4. ### math-calculus

A spherical party balloon is being inflated with helium pumped in at a rate of 12 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 1 ft?

1. ### physics

A 1.6 air bubble is released from the sandy bottom of a warm, shallow sea, where the gauge pressure is 1.6 . The bubble rises slowly enough that the air inside remains at the same constant temperature as the water. What is the

2. ### Calculus: need clarification to where the #'s go

Air is being pumped into a spherical balloon so that its volume increases at a rate of 80 \mbox{cm}^3\mbox{/s}. How fast is the surface area of the balloon increasing when its radius is 14 \mbox{cm}? Recall that a ball of radius r

3. ### Math( Can you help me with starting this problem)

How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dv/dt , the change in volume with respect to time, which is 15 cubic

4. ### calculus

The radius of a spherical balloon is increasing at a rate of 2 centimeters per minute. How fast is the volume changing when the radius is 8 centimeters? Note: The volume of a sphere is given by 4(pi)r^3

1. ### calc

Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 {cm}^3}. How fast is the surface area of the balloon increasing when its radius is 16{cm}?

2. ### physics

Calculate the gauge pressure inside a soap bubble 0.0002m in radius using the surface tension for soapy water

3. ### algebra

Sara goes on a slingshot ride in an amusement park. She is strapped into a spherical ball that has a radius of centimeters. What is the volume of air in the spherical ball?

4. ### science

1) In a syringe, what happens to the air paricles when you push the plunger? 2) In a syring, what happens the the air particles in the bubble from the bubble wrap when you pull up the plunger? 3) Are there more air particles in