Expanding ripple

[Agent Smith]

Is this correct? I don't know what it is, perhaps being the first time solving calculus problems, but something just doesn't feel right about my solution.

Comments/criticisms/approval/etc. welcome

 

[Agent Smith]

@mario99 @khansaheb @fresh_42
Gracias

 

[Agent Smith]

Correction (I think):

$A = \pi r^2$ and $r = 4t$

So $A = \pi (4t)^2$

Let $u = 4t$

$\frac{dA}{dt} = \frac{dA}{du} \times \frac{du}{dt}$

$\frac{dA}{dt} = \frac{d (\pi u^2)}{du} \times \frac{d(4t)}{dt}$

$\frac{dA}{dt} = 2\pi u \times 4 = 8 \pi u = 8 \pi \times 4t = 32 \pi t$

At the 3 m mark, the time is $\frac{3}{4}$ seconds

So at the 3 m mark, $\frac{dA}{dt} = 32 \times \pi \times \frac{3}{4} = 24\pi$ sq. m/second

Correct?

 

[Harry_the_cat]

Both are correct