What is the association between two separate rates of change in finding the derivative of the volume of a sphere?

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Imagine a balloon being inflated. The balloon has a volume $V$ and a radius $r$, which are both functions of time $t$. Finally, suppose $fracdVdt$ is constant.

Is the following fact true, and if so why? (Or why not?)

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdVdrbigg)$$

I understand how $fracdVdr$ represents the rate of change of $V$ with respect to $r$, but not how both quantities, as functions of time, can be multiplied in that manner to produce $fracdVdr$.

Thank you.

derivatives volume

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asked Jul 29 at 20:42

Jamie Corkhill

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up vote
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down vote

favorite

Imagine a balloon being inflated. The balloon has a volume $V$ and a radius $r$, which are both functions of time $t$. Finally, suppose $fracdVdt$ is constant.

Is the following fact true, and if so why? (Or why not?)

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdVdrbigg)$$

I understand how $fracdVdr$ represents the rate of change of $V$ with respect to $r$, but not how both quantities, as functions of time, can be multiplied in that manner to produce $fracdVdr$.

Thank you.

derivatives volume

share|cite|improve this question

asked Jul 29 at 20:42

Jamie Corkhill

1158

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Imagine a balloon being inflated. The balloon has a volume $V$ and a radius $r$, which are both functions of time $t$. Finally, suppose $fracdVdt$ is constant.

Is the following fact true, and if so why? (Or why not?)

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdVdrbigg)$$

I understand how $fracdVdr$ represents the rate of change of $V$ with respect to $r$, but not how both quantities, as functions of time, can be multiplied in that manner to produce $fracdVdr$.

Thank you.

derivatives volume

share|cite|improve this question

asked Jul 29 at 20:42

Jamie Corkhill

1158

Imagine a balloon being inflated. The balloon has a volume $V$ and a radius $r$, which are both functions of time $t$. Finally, suppose $fracdVdt$ is constant.

Is the following fact true, and if so why? (Or why not?)

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdVdrbigg)$$

I understand how $fracdVdr$ represents the rate of change of $V$ with respect to $r$, but not how both quantities, as functions of time, can be multiplied in that manner to produce $fracdVdr$.

Thank you.

derivatives volume

share|cite|improve this question

asked Jul 29 at 20:42

Jamie Corkhill

1158

share|cite|improve this question

share|cite|improve this question

asked Jul 29 at 20:42

Jamie Corkhill

1158

asked Jul 29 at 20:42

Jamie Corkhill

1158

asked Jul 29 at 20:42

Jamie Corkhill

1158

1158

add a comment | 

add a comment | 

2 Answers
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accepted

I think what you're confused about is the idea that a seemingly uninvolved intermediary variable can be used to determine a relationship between two other variables. The following example may help clear things up:

Suppose that Bob sells apples for $5$ dollars each. Bob also loves to eat cookies, and suppose a cookie costs $1$ dollar. Then, for each apple Bob sells, he makes $5$ dollars. For each dollar Bob earns, he can buy one cookie. If the number of apples sold is $A$, the number of dollars Bob earns is $D$, and the number of cookies Bob can buy is $C$, then:$$text5 Cookies per Apple=fracdCdA=fracdCdDcdotfracdDdA=text($1$ Cookie per Dollar)$cdot$($5$ Dollars per Apple)$$

Now, let's see what this means in the context of your problem. $V$ is the volume of the balloon, and $r$ is the radius, so $t$ is our "intermediary variable" (much like how "Dollars=$D$" was in the example). Then, $fracdVdr$ is "volume per radius"—in other words, "how much space the balloon takes up based on its radius". $fracdVdt$ is the "volume per time"— in other words, how fast the balloon is growing per time unit (e.g. cubic meters per second). Lastly, $fracdtdr$ is "seconds per radius"— in other words, if the radius grows by $R$ length units, then we have been inflating the balloon for $T$ time units. Bringing it all together, we have
$$fracdVdr=fracdVdtcdotfracdtdr$$

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answered Jul 29 at 22:01

高田航

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up vote
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By chain rule we have

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdtdrbigg)$$

indeed recall that

$$V=frac43 pi r^3 implies fracdVdr =4pi r^2$$

and

$$fracdVdt =4pi r^2fracdrdt$$

share|cite|improve this answer

edited Jul 29 at 20:50

answered Jul 29 at 20:44

gimusi

64.5k73482

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
  – Jamie Corkhill
  Jul 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
  – gimusi
  Jul 29 at 20:58
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

I think what you're confused about is the idea that a seemingly uninvolved intermediary variable can be used to determine a relationship between two other variables. The following example may help clear things up:

Suppose that Bob sells apples for $5$ dollars each. Bob also loves to eat cookies, and suppose a cookie costs $1$ dollar. Then, for each apple Bob sells, he makes $5$ dollars. For each dollar Bob earns, he can buy one cookie. If the number of apples sold is $A$, the number of dollars Bob earns is $D$, and the number of cookies Bob can buy is $C$, then:$$text5 Cookies per Apple=fracdCdA=fracdCdDcdotfracdDdA=text($1$ Cookie per Dollar)$cdot$($5$ Dollars per Apple)$$

Now, let's see what this means in the context of your problem. $V$ is the volume of the balloon, and $r$ is the radius, so $t$ is our "intermediary variable" (much like how "Dollars=$D$" was in the example). Then, $fracdVdr$ is "volume per radius"—in other words, "how much space the balloon takes up based on its radius". $fracdVdt$ is the "volume per time"— in other words, how fast the balloon is growing per time unit (e.g. cubic meters per second). Lastly, $fracdtdr$ is "seconds per radius"— in other words, if the radius grows by $R$ length units, then we have been inflating the balloon for $T$ time units. Bringing it all together, we have
$$fracdVdr=fracdVdtcdotfracdtdr$$

share|cite|improve this answer

answered Jul 29 at 22:01

高田航

1,116318

add a comment | 

up vote
1
down vote



accepted

I think what you're confused about is the idea that a seemingly uninvolved intermediary variable can be used to determine a relationship between two other variables. The following example may help clear things up:

Suppose that Bob sells apples for $5$ dollars each. Bob also loves to eat cookies, and suppose a cookie costs $1$ dollar. Then, for each apple Bob sells, he makes $5$ dollars. For each dollar Bob earns, he can buy one cookie. If the number of apples sold is $A$, the number of dollars Bob earns is $D$, and the number of cookies Bob can buy is $C$, then:$$text5 Cookies per Apple=fracdCdA=fracdCdDcdotfracdDdA=text($1$ Cookie per Dollar)$cdot$($5$ Dollars per Apple)$$

Now, let's see what this means in the context of your problem. $V$ is the volume of the balloon, and $r$ is the radius, so $t$ is our "intermediary variable" (much like how "Dollars=$D$" was in the example). Then, $fracdVdr$ is "volume per radius"—in other words, "how much space the balloon takes up based on its radius". $fracdVdt$ is the "volume per time"— in other words, how fast the balloon is growing per time unit (e.g. cubic meters per second). Lastly, $fracdtdr$ is "seconds per radius"— in other words, if the radius grows by $R$ length units, then we have been inflating the balloon for $T$ time units. Bringing it all together, we have
$$fracdVdr=fracdVdtcdotfracdtdr$$

share|cite|improve this answer

answered Jul 29 at 22:01

高田航

1,116318

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

I think what you're confused about is the idea that a seemingly uninvolved intermediary variable can be used to determine a relationship between two other variables. The following example may help clear things up:

Suppose that Bob sells apples for $5$ dollars each. Bob also loves to eat cookies, and suppose a cookie costs $1$ dollar. Then, for each apple Bob sells, he makes $5$ dollars. For each dollar Bob earns, he can buy one cookie. If the number of apples sold is $A$, the number of dollars Bob earns is $D$, and the number of cookies Bob can buy is $C$, then:$$text5 Cookies per Apple=fracdCdA=fracdCdDcdotfracdDdA=text($1$ Cookie per Dollar)$cdot$($5$ Dollars per Apple)$$

Now, let's see what this means in the context of your problem. $V$ is the volume of the balloon, and $r$ is the radius, so $t$ is our "intermediary variable" (much like how "Dollars=$D$" was in the example). Then, $fracdVdr$ is "volume per radius"—in other words, "how much space the balloon takes up based on its radius". $fracdVdt$ is the "volume per time"— in other words, how fast the balloon is growing per time unit (e.g. cubic meters per second). Lastly, $fracdtdr$ is "seconds per radius"— in other words, if the radius grows by $R$ length units, then we have been inflating the balloon for $T$ time units. Bringing it all together, we have
$$fracdVdr=fracdVdtcdotfracdtdr$$

share|cite|improve this answer

answered Jul 29 at 22:01

高田航

1,116318

I think what you're confused about is the idea that a seemingly uninvolved intermediary variable can be used to determine a relationship between two other variables. The following example may help clear things up:

Suppose that Bob sells apples for $5$ dollars each. Bob also loves to eat cookies, and suppose a cookie costs $1$ dollar. Then, for each apple Bob sells, he makes $5$ dollars. For each dollar Bob earns, he can buy one cookie. If the number of apples sold is $A$, the number of dollars Bob earns is $D$, and the number of cookies Bob can buy is $C$, then:$$text5 Cookies per Apple=fracdCdA=fracdCdDcdotfracdDdA=text($1$ Cookie per Dollar)$cdot$($5$ Dollars per Apple)$$

Now, let's see what this means in the context of your problem. $V$ is the volume of the balloon, and $r$ is the radius, so $t$ is our "intermediary variable" (much like how "Dollars=$D$" was in the example). Then, $fracdVdr$ is "volume per radius"—in other words, "how much space the balloon takes up based on its radius". $fracdVdt$ is the "volume per time"— in other words, how fast the balloon is growing per time unit (e.g. cubic meters per second). Lastly, $fracdtdr$ is "seconds per radius"— in other words, if the radius grows by $R$ length units, then we have been inflating the balloon for $T$ time units. Bringing it all together, we have
$$fracdVdr=fracdVdtcdotfracdtdr$$

share|cite|improve this answer

answered Jul 29 at 22:01

高田航

1,116318

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 22:01

高田航

1,116318

answered Jul 29 at 22:01

高田航

1,116318

answered Jul 29 at 22:01

高田航

1,116318

1,116318

add a comment | 

add a comment | 

up vote
0
down vote

By chain rule we have

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdtdrbigg)$$

indeed recall that

$$V=frac43 pi r^3 implies fracdVdr =4pi r^2$$

and

$$fracdVdt =4pi r^2fracdrdt$$

share|cite|improve this answer

edited Jul 29 at 20:50

answered Jul 29 at 20:44

gimusi

64.5k73482

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
  – Jamie Corkhill
  Jul 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
  – gimusi
  Jul 29 at 20:58
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

By chain rule we have

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdtdrbigg)$$

indeed recall that

$$V=frac43 pi r^3 implies fracdVdr =4pi r^2$$

and

$$fracdVdt =4pi r^2fracdrdt$$

share|cite|improve this answer

edited Jul 29 at 20:50

answered Jul 29 at 20:44

gimusi

64.5k73482

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
  – Jamie Corkhill
  Jul 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
  – gimusi
  Jul 29 at 20:58
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

By chain rule we have

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdtdrbigg)$$

indeed recall that

$$V=frac43 pi r^3 implies fracdVdr =4pi r^2$$

and

$$fracdVdt =4pi r^2fracdrdt$$

share|cite|improve this answer

edited Jul 29 at 20:50

answered Jul 29 at 20:44

gimusi

64.5k73482

By chain rule we have

$$fracdVdr = bigg(fracdVdtbigg)bigg(fracdtdrbigg)$$

indeed recall that

$$V=frac43 pi r^3 implies fracdVdr =4pi r^2$$

and

$$fracdVdt =4pi r^2fracdrdt$$

share|cite|improve this answer

edited Jul 29 at 20:50

answered Jul 29 at 20:44

gimusi

64.5k73482

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 20:50

edited Jul 29 at 20:50

edited Jul 29 at 20:50

answered Jul 29 at 20:44

gimusi

64.5k73482

answered Jul 29 at 20:44

gimusi

64.5k73482

answered Jul 29 at 20:44

gimusi

64.5k73482

64.5k73482

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
  – Jamie Corkhill
  Jul 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
  – gimusi
  Jul 29 at 20:58
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
  – Jamie Corkhill
  Jul 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
  – gimusi
  Jul 29 at 20:58
  
  

  
  
  
  

Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
– Jamie Corkhill
Jul 29 at 20:55

Thanks. I still have a little bit of trouble understanding how the derivative of two quantities, with respect to different variables, yields the derivative of the volume with respect to the radius.
– Jamie Corkhill
Jul 29 at 20:55

1

1

@JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
– gimusi
Jul 29 at 20:58

@JamieCorkhill In other words we have $$fracdVdt = fracdVdrfracdrdt implies fracdVdr = fracdVdtfracdtdr$$ try with some concrete example.
– gimusi
Jul 29 at 20:58

add a comment | 

 

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