calc help: rate of change w/ respect to t of the surface area
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As a spherical balloon is being inflated, its radius (in centimeters) after $t$ minutes is given by $r(t)=3left[(t+8)^1/3right]$ where $0<t<10$. What is the rate of change with respect to $t$ of each of the following at $t=8$?
- $r(t)$
- the surface area
I solved 1 by taking the derivative and plugging in $t$, so I approached 2 similarly. However, plugging in $8$ into the $r(t)$ equation and plugging that into the surface area equation doesn't seem to be working.
Thank you so much!
calculus surfaces spheres
edited Aug 10 at 22:37
Robert Howard
1,331620
asked Aug 10 at 22:28
r squared
43
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As a spherical balloon is being inflated, its radius (in centimeters) after $t$ minutes is given by $r(t)=3left[(t+8)^1/3right]$ where $0<t<10$. What is the rate of change with respect to $t$ of each of the following at $t=8$?
- $r(t)$
- the surface area
I solved 1 by taking the derivative and plugging in $t$, so I approached 2 similarly. However, plugging in $8$ into the $r(t)$ equation and plugging that into the surface area equation doesn't seem to be working.
Thank you so much!
calculus surfaces spheres
edited Aug 10 at 22:37
Robert Howard
1,331620
asked Aug 10 at 22:28
r squared
43
1
Welcome to MSE! We use something called MathJax here to format mathematics so equations and expressions are easier to read. That's what I did in my edit to your question, and you can find a tutorial to learn how to do it yourself here: math.meta.stackexchange.com/questions/5020/…. It's really easy to pick up, trust me.
– Robert Howard
Aug 10 at 22:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As a spherical balloon is being inflated, its radius (in centimeters) after $t$ minutes is given by $r(t)=3left[(t+8)^1/3right]$ where $0<t<10$. What is the rate of change with respect to $t$ of each of the following at $t=8$?
- $r(t)$
- the surface area
I solved 1 by taking the derivative and plugging in $t$, so I approached 2 similarly. However, plugging in $8$ into the $r(t)$ equation and plugging that into the surface area equation doesn't seem to be working.
Thank you so much!
calculus surfaces spheres
edited Aug 10 at 22:37
Robert Howard
1,331620
asked Aug 10 at 22:28
r squared
43
As a spherical balloon is being inflated, its radius (in centimeters) after $t$ minutes is given by $r(t)=3left[(t+8)^1/3right]$ where $0<t<10$. What is the rate of change with respect to $t$ of each of the following at $t=8$?
- $r(t)$
- the surface area
I solved 1 by taking the derivative and plugging in $t$, so I approached 2 similarly. However, plugging in $8$ into the $r(t)$ equation and plugging that into the surface area equation doesn't seem to be working.
Thank you so much!
calculus surfaces spheres
edited Aug 10 at 22:37
Robert Howard
1,331620
asked Aug 10 at 22:28
r squared
43
edited Aug 10 at 22:37
Robert Howard
1,331620
edited Aug 10 at 22:37
Robert Howard
1,331620
edited Aug 10 at 22:37
Robert Howard
1,331620
1,331620
asked Aug 10 at 22:28
r squared
43
asked Aug 10 at 22:28
r squared
43
asked Aug 10 at 22:28
r squared
43
43
1
Welcome to MSE! We use something called MathJax here to format mathematics so equations and expressions are easier to read. That's what I did in my edit to your question, and you can find a tutorial to learn how to do it yourself here: math.meta.stackexchange.com/questions/5020/…. It's really easy to pick up, trust me.
– Robert Howard
Aug 10 at 22:39
add a comment |Â
1
Welcome to MSE! We use something called MathJax here to format mathematics so equations and expressions are easier to read. That's what I did in my edit to your question, and you can find a tutorial to learn how to do it yourself here: math.meta.stackexchange.com/questions/5020/…. It's really easy to pick up, trust me.
– Robert Howard
Aug 10 at 22:39
1
1
Welcome to MSE! We use something called MathJax here to format mathematics so equations and expressions are easier to read. That's what I did in my edit to your question, and you can find a tutorial to learn how to do it yourself here: math.meta.stackexchange.com/questions/5020/…. It's really easy to pick up, trust me.
– Robert Howard
Aug 10 at 22:39
Welcome to MSE! We use something called MathJax here to format mathematics so equations and expressions are easier to read. That's what I did in my edit to your question, and you can find a tutorial to learn how to do it yourself here: math.meta.stackexchange.com/questions/5020/…. It's really easy to pick up, trust me.
– Robert Howard
Aug 10 at 22:39
add a comment |Â
1 Answer
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1
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You cannot plug in the derivative of $r(t)$ into the expression for the surface area. You want the derivative of $A(t)$, where $A$ is the surface area.
The equation of the surface area of a sphere is $$A = 4pi r^2$$ So, plugging in the expression for $r$, $$A = 4pi (3(t+8)^frac13)^2$$$$A = 4pi (9(t+8)^frac23)$$$$A = 36pi (t+8)^frac23$$
Now take the derivative of that and plug in $t=8.$
answered Aug 10 at 22:38
RayDansh
873215
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You cannot plug in the derivative of $r(t)$ into the expression for the surface area. You want the derivative of $A(t)$, where $A$ is the surface area.
The equation of the surface area of a sphere is $$A = 4pi r^2$$ So, plugging in the expression for $r$, $$A = 4pi (3(t+8)^frac13)^2$$$$A = 4pi (9(t+8)^frac23)$$$$A = 36pi (t+8)^frac23$$
Now take the derivative of that and plug in $t=8.$
answered Aug 10 at 22:38
RayDansh
873215
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
add a comment |Â
up vote
1
down vote
You cannot plug in the derivative of $r(t)$ into the expression for the surface area. You want the derivative of $A(t)$, where $A$ is the surface area.
The equation of the surface area of a sphere is $$A = 4pi r^2$$ So, plugging in the expression for $r$, $$A = 4pi (3(t+8)^frac13)^2$$$$A = 4pi (9(t+8)^frac23)$$$$A = 36pi (t+8)^frac23$$
Now take the derivative of that and plug in $t=8.$
answered Aug 10 at 22:38
RayDansh
873215
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You cannot plug in the derivative of $r(t)$ into the expression for the surface area. You want the derivative of $A(t)$, where $A$ is the surface area.
The equation of the surface area of a sphere is $$A = 4pi r^2$$ So, plugging in the expression for $r$, $$A = 4pi (3(t+8)^frac13)^2$$$$A = 4pi (9(t+8)^frac23)$$$$A = 36pi (t+8)^frac23$$
Now take the derivative of that and plug in $t=8.$
answered Aug 10 at 22:38
RayDansh
873215
You cannot plug in the derivative of $r(t)$ into the expression for the surface area. You want the derivative of $A(t)$, where $A$ is the surface area.
The equation of the surface area of a sphere is $$A = 4pi r^2$$ So, plugging in the expression for $r$, $$A = 4pi (3(t+8)^frac13)^2$$$$A = 4pi (9(t+8)^frac23)$$$$A = 36pi (t+8)^frac23$$
Now take the derivative of that and plug in $t=8.$
answered Aug 10 at 22:38
RayDansh
873215
answered Aug 10 at 22:38
RayDansh
873215
answered Aug 10 at 22:38
RayDansh
873215
answered Aug 10 at 22:38
RayDansh
873215
873215
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
add a comment |Â
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
I'm extremely sorry for not using mathjax (gotta learn that), but after doing that I have 24(pi)[(16)^-1/3], which is very different from the textbook solution of (6)(pi)4^1/3. Am I calculating the derivative wrong? thanks, again.
– r squared
Aug 10 at 22:44
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
You did the differentiation correctly. While written completely differently, those two answers are exactly the same - around $9.524pi$.
– RayDansh
Aug 10 at 22:46
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
Wow, I feel ridiculously stupid. Thanks, again @RayDansh
– r squared
Aug 10 at 23:09
add a comment |Â
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1
Welcome to MSE! We use something called MathJax here to format mathematics so equations and expressions are easier to read. That's what I did in my edit to your question, and you can find a tutorial to learn how to do it yourself here: math.meta.stackexchange.com/questions/5020/…. It's really easy to pick up, trust me.
– Robert Howard
Aug 10 at 22:39