Related rates, Differential problem. What is wrong with my work?

Question

Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm^2?

My work:

Given = 6 cm/s
Unknown= dx/dt when x = 4

A=x^2

(dA/dr)(dr/dt)=2x(dx/dt)

(dA/dt)=2x(dx/dt)

6 cm/s=2(4)cm(dx/dt)

6/8 s=dx/dt


The obvious thing i have wrong is that the rate turned into seconds. The second thing is, the answer is 48cm^2/s.

Please help! I already tried doing wierd things to the problem like using 36cm^2/s^2 and 96cm^3/s as the rates. . .

Answer 1

dx/dt is not the unknown bcause it is the increasing rate w/c is 6 cm/s.
dA/dt is the unknown.

x = 16 cm^2
x=4 cm
dx/dt = 6 cm/s
dA/dt = ?

A = x^2

differentiating it, we have
dA/dt = 2x dx/dt

substituting the values,
dA/dt = 2(4 cm) 6cm/s

dA/dt = 48 cm^2/s is the answer!

Answer 2

Greetings,

6 cm/s is dx/dt

Then dA/dt = 2*4*6 =48

Regards

Answer 3

It helps to define your terms. As it is, you're confusing yourself. I see both r and x. You only need x.

Let
x = length of side
A = area
t = time in seconds

Given
dx/dt = 6 cm/sec

Find
dA/dt
when A = 16 cm²
____________

We have

A = x² = 16
x = 4 cm

dA/dx = 2x

dA/dt = (dA/dx)(dx/dt) = 2x(6) = 12x
dA/dt = 12*4 = 48 cm²/sec

when A = 16 cm²

Answer 4

6 cm/s is dx/dt

Then dA/dt = 2*4*6 =48

Answer 5

rate is the area of the square - dS/dt

S = x^2 so

dS/dt = 2*x(dx/dt) = 2*x*6 cm/s

when s =16 x = 4 so

dS/dt = 2*4*6 cm^2/s = 48 cm^2/s

Answer 6

you are looking for rate of change of area (dA/dt)...

A = x^2; when A=16, x = 4

differentiate then substitute

dA/dt = 2x(dx/dt)

= 2(4cm)(6cm/s) =48 cm^2/s

Answer 7

<> To be honest to Trevor, he did say the very similar element. inspect his fourth paragraph. the completed communicate grew to change into slowed down with the tree difficulty, which looked fairly tangential to me, so i can see it may be uncomplicated to omit, yet a more suitable careful examining ought to forestall a number of those stupid questions contained in the destiny. there is little question that, surely, the project Dook poses -- the problem of the time lag -- calls for differential equations to respond to, and the precise cost ought to matter on dCO2/dt and the complexity of your variety. i don't believe of it should be too complicated to workout consultation a crude approximation utilizing some thing like an uncomplicated container variety (which, for sure, includes a equipment of differential equations). i'm no longer effective it ought to furnish a lot illumination to a majority of posters the following, it truly is why a more suitable qualitative answer often suffices, yet I agree that it truly is disingenuous of Trevor to intend that the project -- which, finally, he did not fairly address -- would nicely be solved without DEs. Dook <> on condition that the WV comments is a function of temperature, they're an similar question. in truth, we do not go with *any* advice about the comments itself (e.g. Clausius-Clapeyron), except the mere incontrovertible truth that it responds to temperature, to respond to your question.