If a circle's area gets increased at the rate of 2 $cm^2. s^-1$ .Then what will be the increase rate of it's radius after $frac2811$ seconds?
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If a circle's area gets increased at the rate of 2 $cm^2. s^-1$ .Then what will be the increase rate of it's radius after $frac2811$ seconds?
My Try : $fracdsdt= 2pi r fracdrdt$ so we can write $2pi r fracdrdt = 2$. Can anyone help me out with this problem?
calculus derivatives
edited Aug 31 at 10:35
MRobinson
53914
asked Aug 31 at 7:37
cmi
936110
add a comment |Â
up vote
1
down vote
favorite
1
If a circle's area gets increased at the rate of 2 $cm^2. s^-1$ .Then what will be the increase rate of it's radius after $frac2811$ seconds?
My Try : $fracdsdt= 2pi r fracdrdt$ so we can write $2pi r fracdrdt = 2$. Can anyone help me out with this problem?
calculus derivatives
edited Aug 31 at 10:35
MRobinson
53914
asked Aug 31 at 7:37
cmi
936110
What dimension of the circle is being increased? (and the rate needs to be per second).
– MRobinson
Aug 31 at 7:42
1
@MRobinson I think the OP is referring to increase in area. Also, I have corrected the unit (edit pending review).
– Devashish Kaushik
Aug 31 at 7:44
@DevashishKaushik of course, I hadn't noticed the $cm^2$! Thank you
– MRobinson
Aug 31 at 7:51
add a comment |Â
up vote
1
down vote
favorite
1
up vote
1
down vote
favorite
1
1
If a circle's area gets increased at the rate of 2 $cm^2. s^-1$ .Then what will be the increase rate of it's radius after $frac2811$ seconds?
My Try : $fracdsdt= 2pi r fracdrdt$ so we can write $2pi r fracdrdt = 2$. Can anyone help me out with this problem?
calculus derivatives
edited Aug 31 at 10:35
MRobinson
53914
asked Aug 31 at 7:37
cmi
936110
If a circle's area gets increased at the rate of 2 $cm^2. s^-1$ .Then what will be the increase rate of it's radius after $frac2811$ seconds?
My Try : $fracdsdt= 2pi r fracdrdt$ so we can write $2pi r fracdrdt = 2$. Can anyone help me out with this problem?
calculus derivatives
calculus derivatives
edited Aug 31 at 10:35
MRobinson
53914
asked Aug 31 at 7:37
cmi
936110
edited Aug 31 at 10:35
MRobinson
53914
asked Aug 31 at 7:37
cmi
936110
edited Aug 31 at 10:35
MRobinson
53914
edited Aug 31 at 10:35
MRobinson
53914
edited Aug 31 at 10:35
MRobinson
53914
53914
asked Aug 31 at 7:37
cmi
936110
asked Aug 31 at 7:37
cmi
936110
asked Aug 31 at 7:37
cmi
936110
936110
What dimension of the circle is being increased? (and the rate needs to be per second).
– MRobinson
Aug 31 at 7:42
1
@MRobinson I think the OP is referring to increase in area. Also, I have corrected the unit (edit pending review).
– Devashish Kaushik
Aug 31 at 7:44
@DevashishKaushik of course, I hadn't noticed the $cm^2$! Thank you
– MRobinson
Aug 31 at 7:51
add a comment |Â
What dimension of the circle is being increased? (and the rate needs to be per second).
– MRobinson
Aug 31 at 7:42
1
@MRobinson I think the OP is referring to increase in area. Also, I have corrected the unit (edit pending review).
– Devashish Kaushik
Aug 31 at 7:44
@DevashishKaushik of course, I hadn't noticed the $cm^2$! Thank you
– MRobinson
Aug 31 at 7:51
What dimension of the circle is being increased? (and the rate needs to be per second).
– MRobinson
Aug 31 at 7:42
What dimension of the circle is being increased? (and the rate needs to be per second).
– MRobinson
Aug 31 at 7:42
1
1
@MRobinson I think the OP is referring to increase in area. Also, I have corrected the unit (edit pending review).
– Devashish Kaushik
Aug 31 at 7:44
@MRobinson I think the OP is referring to increase in area. Also, I have corrected the unit (edit pending review).
– Devashish Kaushik
Aug 31 at 7:44
@DevashishKaushik of course, I hadn't noticed the $cm^2$! Thank you
– MRobinson
Aug 31 at 7:51
@DevashishKaushik of course, I hadn't noticed the $cm^2$! Thank you
– MRobinson
Aug 31 at 7:51
add a comment |Â
1 Answer
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up vote
1
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Yes your way is correct indeed we have
$$A(t)=pi r(t)^2implies A'(t)=2 pi r(t)r'(t)=2 implies r'(t)=frac1pi r(t)$$
and therefore
$$fracdrdt=frac1pi r implies rdr=frac 1 pi dt implies frac12r^2=frac t pi+C implies r(t)=sqrtfrac2tpi+C$$
and assuming $r(0)=r_0 implies C=r_0^2$ that is
$$r'(t)=frac1sqrt2pi t+pi r_0^2$$
answered Aug 31 at 7:54
gimusi
71.7k73787
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes your way is correct indeed we have
$$A(t)=pi r(t)^2implies A'(t)=2 pi r(t)r'(t)=2 implies r'(t)=frac1pi r(t)$$
and therefore
$$fracdrdt=frac1pi r implies rdr=frac 1 pi dt implies frac12r^2=frac t pi+C implies r(t)=sqrtfrac2tpi+C$$
and assuming $r(0)=r_0 implies C=r_0^2$ that is
$$r'(t)=frac1sqrt2pi t+pi r_0^2$$
answered Aug 31 at 7:54
gimusi
71.7k73787
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
 |Â
show 1 more comment
up vote
1
down vote
accepted
Yes your way is correct indeed we have
$$A(t)=pi r(t)^2implies A'(t)=2 pi r(t)r'(t)=2 implies r'(t)=frac1pi r(t)$$
and therefore
$$fracdrdt=frac1pi r implies rdr=frac 1 pi dt implies frac12r^2=frac t pi+C implies r(t)=sqrtfrac2tpi+C$$
and assuming $r(0)=r_0 implies C=r_0^2$ that is
$$r'(t)=frac1sqrt2pi t+pi r_0^2$$
answered Aug 31 at 7:54
gimusi
71.7k73787
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes your way is correct indeed we have
$$A(t)=pi r(t)^2implies A'(t)=2 pi r(t)r'(t)=2 implies r'(t)=frac1pi r(t)$$
and therefore
$$fracdrdt=frac1pi r implies rdr=frac 1 pi dt implies frac12r^2=frac t pi+C implies r(t)=sqrtfrac2tpi+C$$
and assuming $r(0)=r_0 implies C=r_0^2$ that is
$$r'(t)=frac1sqrt2pi t+pi r_0^2$$
answered Aug 31 at 7:54
gimusi
71.7k73787
Yes your way is correct indeed we have
$$A(t)=pi r(t)^2implies A'(t)=2 pi r(t)r'(t)=2 implies r'(t)=frac1pi r(t)$$
and therefore
$$fracdrdt=frac1pi r implies rdr=frac 1 pi dt implies frac12r^2=frac t pi+C implies r(t)=sqrtfrac2tpi+C$$
and assuming $r(0)=r_0 implies C=r_0^2$ that is
$$r'(t)=frac1sqrt2pi t+pi r_0^2$$
answered Aug 31 at 7:54
gimusi
71.7k73787
edited Aug 31 at 8:49
answered Aug 31 at 7:54
gimusi
71.7k73787
answered Aug 31 at 7:54
gimusi
71.7k73787
answered Aug 31 at 7:54
gimusi
71.7k73787
71.7k73787
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
 |Â
show 1 more comment
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@gimsu The OP had asked for increase rate $dr/dt$ of the radius , not the radius itself. It might be norw useful to point out that he just had to plug in the given values in the expression he reached, instead of repeating the calculation and finding the radius. ( No offence intended ) :-)
– Devashish Kaushik
Aug 31 at 7:59
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
@DevashishKaushik I get your point but as you can see the rate $r'(t)$ depends upon $r(t)$ therefore we need to find $r(t)$ to solve the problem completely. For example assuming $r(0)=0$ we can explicitely find a value for the rate at $t=28/11 s$.
– gimusi
Aug 31 at 8:00
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
Of course, I was referring to the fact that seeing that the OP reached $drover dt = f(R)$ , it might be better to give him a slight hint instead of the succeeding steps in their entirety. I am sorry if it seemed like I was suggesting that your solution was incorrect.
– Devashish Kaushik
Aug 31 at 8:22
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
@DevashishKaushik I didn't have that idea about your kind observation. Thanks
– gimusi
Aug 31 at 8:29
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
I think $r(0) = r_0$@gimusi
– cmi
Aug 31 at 8:46
 |Â
show 1 more comment
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What dimension of the circle is being increased? (and the rate needs to be per second).
– MRobinson
Aug 31 at 7:42
1
@MRobinson I think the OP is referring to increase in area. Also, I have corrected the unit (edit pending review).
– Devashish Kaushik
Aug 31 at 7:44
@DevashishKaushik of course, I hadn't noticed the $cm^2$! Thank you
– MRobinson
Aug 31 at 7:51