What is the Volume Change Rate of a Spherical Balloon?

In summary: So, you would need to solve for V'(t), V''(t), and V'''(t). In summary, the volume of a spherical balloon changes with the radius, and the rate at which the volume changes with respect to radius is proportional to the fourth power of the radius. The volume of the balloon increases by approximately 3.14 times when the radius changes from 2 to 2.2ft.
  • #1
jtt
16
0

Homework Statement


the volume v=(4/3)(pie symbol 3.14..)r of a spherical balloon changes with the radius

a)at what rate does the volume change with respect to radius when r= 2ft?
b) by approximately how much does the volume increase when the radius changes from 2 to 2.2ft?

Homework Equations


v=(4/3)(pie symbol 3.14..)r
sorry, couldn't find the pie symbol to multiply the r with in this equation


The Attempt at a Solution

for part a i tried to plug in 2 or the r in the equation and then divide by 2
for part b i tried subtracting the equation when r is substituted for 2 from when the r is substituted for 2.2
 
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  • #2
jtt said:

Homework Statement


the volume v=(4/3)(pie symbol 3.14..)r of a spherical balloon changes with the radius

a)at what rate does the volume change with respect to radius when r= 2ft?
b) by approximately how much does the volume increase when the radius changes from 2 to 2.2ft?

Homework Equations


v=(4/3)(pie symbol 3.14..)r
sorry, couldn't find the pie symbol to multiply the r with in this equation


The Attempt at a Solution

for part a i tried to plug in 2 or the r in the equation and then divide by 2
for part b i tried subtracting the equation when r is substituted for 2 from when the r is substituted for 2.2
Your textbook should have some examples of this type of problem. Take a look at the examples, and then make a more serious effort at solving this one.
 
  • #3
Dear Mark44,
the textbook that i have does not give an example of the volume of a spherical balloon. I'm sorry if it seems as though i am not making a serious attempt at solving this problem but sometimes i have a hard time trying to determine what i need to do to solve math problems, there fore making me a seem as though i am learning at a slower pace than most people or at least in my class. i don't mean to sound lazy and is just the type of person who wants someone to do all the work for me. its just that if i struggle on my own for too long and i feel as though I'm not getting any where, that is when i determine that i need help. please don't take offense if any of my attempts to solve the problem seem lame or that I'm not taking the problem seriously enough. my teacher has high expectations for each individual student to do his or her best no matter what and that i always plan to do, even if nothing i do is correct the first time around. please don't take offense to this if you feel as though the tone of this reply is meant to do just that, i am just one of many frustrated and over worked high school students hoping to pass calculus with a passing grade because calculus, geometry, physics or any other type of math or science is just not what i can comprehend as fast as others. although based on how i wrote this reply you might infer that I'm my strengths is stronger in reading and writing.

if you, Mark44, or anyone else can help me in solving the volume of a spherical balloon, i will be grateful for you taking the time trying to help me get through this
 
  • #4
Your book might not have an example that is just like this problem, but it probably has examples that are similar enough to help you with this problem.

In related rates problems you generally need to establish two equations. One equation represents the relationship between basic quantities in the problem. The other equation represents the rates of change of the basic quantities (i.e., the derivatives with respect to time).

Your first equation is the relationship between the volume V of the balloon and its radius, and that is V = (4/3)[itex]\pi[/itex]r3.
For your second equation, realize that as time goes by, V is changing and r is changing. These should suggest derivatives.
 

Related to What is the Volume Change Rate of a Spherical Balloon?

What is the formula for finding the volume of a spherical balloon?

The formula for finding the volume of a spherical balloon is V = (4/3)πr³, where V is the volume and r is the radius of the balloon.

How do I measure the radius of a spherical balloon?

The radius of a spherical balloon can be measured by using a ruler or measuring tape to measure the distance from the center of the balloon to the edge.

What units should be used when calculating the volume of a spherical balloon?

The units used for calculating the volume of a spherical balloon should be in cubic units, such as cubic centimeters (cm³) or cubic inches (in³).

Can the volume of a spherical balloon change?

Yes, the volume of a spherical balloon can change depending on the amount of air or gas that is pumped into it. The more air or gas that is added, the larger the volume will be.

How is the volume of a spherical balloon affected by changes in temperature?

The volume of a spherical balloon is affected by changes in temperature because temperature can cause the air or gas inside the balloon to expand or contract. As a result, the volume of the balloon will increase or decrease accordingly.

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