Mixing
How fast is the area of rectangle increasing?
$begingroup$
The length of a rectangle is increasing at a rate of 8 cm/s and
its width is increasing at a rate of $3$ cm/s . When the length is
20 cm and the width is 10 cm, how fast is the area of the rectangle
increasing?
So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.
Which I found on the int:
$A=lw$ then take derivative $frac{dA}{dt}= frac{dl}{dt}.w + l.frac{dw}{dt}$
using given number $frac{dA}{dt}= (8)(10) + (20)(3)$
My answer: given numbers--> $frac{dl}{dt}= 8$, $frac{dw}{dt}=3$, $l=20$, $w =10$
so $A=wl$ when I wanna write $w$ in terms of $l$ ----> $l=2w$
so $A=2w*w$ when I take derivative of it ---> $frac{dA}{dw}= 4w $
according to chain rule $frac{dA}{dt}= frac{dA}{dw}frac{dw}{dt}$
when I put the numbers ----> $4w*3$ and we know that $w=10$
It should be 120. I think I found my mistake but still couldn't understand why. I write $w$ in terms of l but if I do the other way then the result is 160. What am I doing wrong?
calculus
edited Jan 17 at 20:57
user144410
1,0282719
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
$endgroup$
|
show 6 more comments
$begingroup$
The length of a rectangle is increasing at a rate of 8 cm/s and
its width is increasing at a rate of $3$ cm/s . When the length is
20 cm and the width is 10 cm, how fast is the area of the rectangle
increasing?
So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.
Which I found on the int:
$A=lw$ then take derivative $frac{dA}{dt}= frac{dl}{dt}.w + l.frac{dw}{dt}$
using given number $frac{dA}{dt}= (8)(10) + (20)(3)$
My answer: given numbers--> $frac{dl}{dt}= 8$, $frac{dw}{dt}=3$, $l=20$, $w =10$
so $A=wl$ when I wanna write $w$ in terms of $l$ ----> $l=2w$
so $A=2w*w$ when I take derivative of it ---> $frac{dA}{dw}= 4w $
according to chain rule $frac{dA}{dt}= frac{dA}{dw}frac{dw}{dt}$
when I put the numbers ----> $4w*3$ and we know that $w=10$
It should be 120. I think I found my mistake but still couldn't understand why. I write $w$ in terms of l but if I do the other way then the result is 160. What am I doing wrong?
calculus
edited Jan 17 at 20:57
user144410
1,0282719
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
$endgroup$
$begingroup$
Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something?
$endgroup$
– Yanko
Jan 17 at 20:59
3
$begingroup$
$l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values.
$endgroup$
– Matteo
Jan 17 at 21:01
$begingroup$
Your expression for $frac{dA}{dt}$ is correct. It should be $140$.
$endgroup$
– John Douma
Jan 17 at 21:03
$begingroup$
@Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– Matteo
Jan 17 at 21:10
$begingroup$
@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
$endgroup$
– fleablood
Jan 17 at 21:22
|
show 6 more comments
$begingroup$
The length of a rectangle is increasing at a rate of 8 cm/s and
its width is increasing at a rate of $3$ cm/s . When the length is
20 cm and the width is 10 cm, how fast is the area of the rectangle
increasing?
So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.
Which I found on the int:
$A=lw$ then take derivative $frac{dA}{dt}= frac{dl}{dt}.w + l.frac{dw}{dt}$
using given number $frac{dA}{dt}= (8)(10) + (20)(3)$
My answer: given numbers--> $frac{dl}{dt}= 8$, $frac{dw}{dt}=3$, $l=20$, $w =10$
so $A=wl$ when I wanna write $w$ in terms of $l$ ----> $l=2w$
so $A=2w*w$ when I take derivative of it ---> $frac{dA}{dw}= 4w $
according to chain rule $frac{dA}{dt}= frac{dA}{dw}frac{dw}{dt}$
when I put the numbers ----> $4w*3$ and we know that $w=10$
It should be 120. I think I found my mistake but still couldn't understand why. I write $w$ in terms of l but if I do the other way then the result is 160. What am I doing wrong?
calculus
edited Jan 17 at 20:57
user144410
1,0282719
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
$endgroup$
The length of a rectangle is increasing at a rate of 8 cm/s and
its width is increasing at a rate of $3$ cm/s . When the length is
20 cm and the width is 10 cm, how fast is the area of the rectangle
increasing?
So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.
Which I found on the int:
$A=lw$ then take derivative $frac{dA}{dt}= frac{dl}{dt}.w + l.frac{dw}{dt}$
using given number $frac{dA}{dt}= (8)(10) + (20)(3)$
My answer: given numbers--> $frac{dl}{dt}= 8$, $frac{dw}{dt}=3$, $l=20$, $w =10$
so $A=wl$ when I wanna write $w$ in terms of $l$ ----> $l=2w$
so $A=2w*w$ when I take derivative of it ---> $frac{dA}{dw}= 4w $
according to chain rule $frac{dA}{dt}= frac{dA}{dw}frac{dw}{dt}$
when I put the numbers ----> $4w*3$ and we know that $w=10$
It should be 120. I think I found my mistake but still couldn't understand why. I write $w$ in terms of l but if I do the other way then the result is 160. What am I doing wrong?
calculus
calculus
edited Jan 17 at 20:57
user144410
1,0282719
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
edited Jan 17 at 20:57
user144410
1,0282719
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
edited Jan 17 at 20:57
user144410
1,0282719
edited Jan 17 at 20:57
user144410
1,0282719
edited Jan 17 at 20:57
user144410
1,0282719
1,0282719
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
asked Jan 17 at 20:40
DisintegratorsDisintegrators
161
161
$begingroup$
Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something?
$endgroup$
– Yanko
Jan 17 at 20:59
3
$begingroup$
$l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values.
$endgroup$
– Matteo
Jan 17 at 21:01
$begingroup$
Your expression for $frac{dA}{dt}$ is correct. It should be $140$.
$endgroup$
– John Douma
Jan 17 at 21:03
$begingroup$
@Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– Matteo
Jan 17 at 21:10
$begingroup$
@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
$endgroup$
– fleablood
Jan 17 at 21:22
|
show 6 more comments
$begingroup$
Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something?
$endgroup$
– Yanko
Jan 17 at 20:59
3
$begingroup$
$l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values.
$endgroup$
– Matteo
Jan 17 at 21:01
$begingroup$
Your expression for $frac{dA}{dt}$ is correct. It should be $140$.
$endgroup$
– John Douma
Jan 17 at 21:03
$begingroup$
@Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– Matteo
Jan 17 at 21:10
$begingroup$
@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
$endgroup$
– fleablood
Jan 17 at 21:22
$begingroup$
Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something?
$endgroup$
– Yanko
Jan 17 at 20:59
$begingroup$
Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something?
$endgroup$
– Yanko
Jan 17 at 20:59
3
3
$begingroup$
$l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values.
$endgroup$
– Matteo
Jan 17 at 21:01
$begingroup$
$l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values.
$endgroup$
– Matteo
Jan 17 at 21:01
$begingroup$
Your expression for $frac{dA}{dt}$ is correct. It should be $140$.
$endgroup$
– John Douma
Jan 17 at 21:03
$begingroup$
Your expression for $frac{dA}{dt}$ is correct. It should be $140$.
$endgroup$
– John Douma
Jan 17 at 21:03
$begingroup$
@Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– Matteo
Jan 17 at 21:10
$begingroup$
@Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– Matteo
Jan 17 at 21:10
$begingroup$
@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
$endgroup$
– fleablood
Jan 17 at 21:22
$begingroup$
@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
$endgroup$
– fleablood
Jan 17 at 21:22
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.
The rate of increase of the area must be close to
$$frac{20.008cdot10.003-20cdot10}{10^{-3}}=140.024.$$
With one microsecond, we get
$$frac{20.000008cdot10.000003-20cdot10}{10^{-6}}=140.000024.$$
This confirms the answer $140$.
The reason why your method doesn't work is because
$$frac{20}{10}nefrac{8}{3}.$$
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.
If they were then it would be correct to say $A = 4w frac {dw}{dt}$
But as they are changing at different rates, you need to use the chain rule.
Perhaps a visualization will help.
We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.
the green areas sum to $l (dw)$ and red areas $w (dl)$
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
$endgroup$
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
add a comment |
$begingroup$
You need to think of $l$ and $w$ as functions of time.
$l(t) = L + 8t$ where $L$ is the initial length. And $w(t) = W+3t$ and, yes, $frac {dl}{dt}=8$ and $frac {dw}{dt} = 3$ but $l ne 20$ and $wne 10$. (That'd mean they are constant functions. They aren't.) $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.
To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.
Now we just can't say $l(t) = 2w(t)$ because that just isn't true. You could say $ {l(t)} = frac {20+8t}{10+3t}{w(t)}$ and $A(t) = frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated. (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.
The rate of increase of the area must be close to
$$frac{20.008cdot10.003-20cdot10}{10^{-3}}=140.024.$$
With one microsecond, we get
$$frac{20.000008cdot10.000003-20cdot10}{10^{-6}}=140.000024.$$
This confirms the answer $140$.
The reason why your method doesn't work is because
$$frac{20}{10}nefrac{8}{3}.$$
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.
The rate of increase of the area must be close to
$$frac{20.008cdot10.003-20cdot10}{10^{-3}}=140.024.$$
With one microsecond, we get
$$frac{20.000008cdot10.000003-20cdot10}{10^{-6}}=140.000024.$$
This confirms the answer $140$.
The reason why your method doesn't work is because
$$frac{20}{10}nefrac{8}{3}.$$
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.
The rate of increase of the area must be close to
$$frac{20.008cdot10.003-20cdot10}{10^{-3}}=140.024.$$
With one microsecond, we get
$$frac{20.000008cdot10.000003-20cdot10}{10^{-6}}=140.000024.$$
This confirms the answer $140$.
The reason why your method doesn't work is because
$$frac{20}{10}nefrac{8}{3}.$$
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
$endgroup$
One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.
The rate of increase of the area must be close to
$$frac{20.008cdot10.003-20cdot10}{10^{-3}}=140.024.$$
With one microsecond, we get
$$frac{20.000008cdot10.000003-20cdot10}{10^{-6}}=140.000024.$$
This confirms the answer $140$.
The reason why your method doesn't work is because
$$frac{20}{10}nefrac{8}{3}.$$
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
answered Jan 17 at 22:09
Yves DaoustYves Daoust
129k675227
129k675227
add a comment |
add a comment |
$begingroup$
While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.
If they were then it would be correct to say $A = 4w frac {dw}{dt}$
But as they are changing at different rates, you need to use the chain rule.
Perhaps a visualization will help.
We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.
the green areas sum to $l (dw)$ and red areas $w (dl)$
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
$endgroup$
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
add a comment |
$begingroup$
While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.
If they were then it would be correct to say $A = 4w frac {dw}{dt}$
But as they are changing at different rates, you need to use the chain rule.
Perhaps a visualization will help.
We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.
the green areas sum to $l (dw)$ and red areas $w (dl)$
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
$endgroup$
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
add a comment |
$begingroup$
While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.
If they were then it would be correct to say $A = 4w frac {dw}{dt}$
But as they are changing at different rates, you need to use the chain rule.
Perhaps a visualization will help.
We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.
the green areas sum to $l (dw)$ and red areas $w (dl)$
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
$endgroup$
While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.
If they were then it would be correct to say $A = 4w frac {dw}{dt}$
But as they are changing at different rates, you need to use the chain rule.
Perhaps a visualization will help.
We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.
the green areas sum to $l (dw)$ and red areas $w (dl)$
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
edited Jan 17 at 21:21
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
answered Jan 17 at 21:16
Doug MDoug M
45.2k31954
45.2k31954
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
add a comment |
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website.
$endgroup$
– Disintegrators
Jan 17 at 21:55
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
More precisely, the rates shouldn't be equal but proportional to the respective sides.
$endgroup$
– Yves Daoust
Jan 17 at 22:02
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
$begingroup$
@Disintegrators I thought I had. Looks fine on my screen.
$endgroup$
– Doug M
Jan 17 at 22:35
add a comment |
$begingroup$
You need to think of $l$ and $w$ as functions of time.
$l(t) = L + 8t$ where $L$ is the initial length. And $w(t) = W+3t$ and, yes, $frac {dl}{dt}=8$ and $frac {dw}{dt} = 3$ but $l ne 20$ and $wne 10$. (That'd mean they are constant functions. They aren't.) $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.
To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.
Now we just can't say $l(t) = 2w(t)$ because that just isn't true. You could say $ {l(t)} = frac {20+8t}{10+3t}{w(t)}$ and $A(t) = frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated. (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
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$begingroup$
You need to think of $l$ and $w$ as functions of time.
$l(t) = L + 8t$ where $L$ is the initial length. And $w(t) = W+3t$ and, yes, $frac {dl}{dt}=8$ and $frac {dw}{dt} = 3$ but $l ne 20$ and $wne 10$. (That'd mean they are constant functions. They aren't.) $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.
To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.
Now we just can't say $l(t) = 2w(t)$ because that just isn't true. You could say $ {l(t)} = frac {20+8t}{10+3t}{w(t)}$ and $A(t) = frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated. (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
$endgroup$
add a comment |
$begingroup$
You need to think of $l$ and $w$ as functions of time.
$l(t) = L + 8t$ where $L$ is the initial length. And $w(t) = W+3t$ and, yes, $frac {dl}{dt}=8$ and $frac {dw}{dt} = 3$ but $l ne 20$ and $wne 10$. (That'd mean they are constant functions. They aren't.) $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.
To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.
Now we just can't say $l(t) = 2w(t)$ because that just isn't true. You could say $ {l(t)} = frac {20+8t}{10+3t}{w(t)}$ and $A(t) = frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated. (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
$endgroup$
You need to think of $l$ and $w$ as functions of time.
$l(t) = L + 8t$ where $L$ is the initial length. And $w(t) = W+3t$ and, yes, $frac {dl}{dt}=8$ and $frac {dw}{dt} = 3$ but $l ne 20$ and $wne 10$. (That'd mean they are constant functions. They aren't.) $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.
To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.
Now we just can't say $l(t) = 2w(t)$ because that just isn't true. You could say $ {l(t)} = frac {20+8t}{10+3t}{w(t)}$ and $A(t) = frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated. (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
answered Jan 17 at 21:57
fleabloodfleablood
71.5k22686
71.5k22686
add a comment |
add a comment |
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Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something?
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– Yanko
Jan 17 at 20:59
3
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$l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values.
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– Matteo
Jan 17 at 21:01
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Your expression for $frac{dA}{dt}$ is correct. It should be $140$.
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– John Douma
Jan 17 at 21:03
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@Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
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– Matteo
Jan 17 at 21:10
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@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
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– fleablood
Jan 17 at 21:22