Troubleshooting a Simple Problem: Tips & Tricks

[karush]

View attachment 1554

I know this is a simple problem but new at it. answer not in book so hope correct.

 

[MarkFL]

Re: expanding cube

You have the formula for volume:

\(\displaystyle V=s^3\)

Differentiating with respect to time $t$, we find:

\(\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}\)

Now plug in the given data...and keep in mind your units...you should get \(\displaystyle \frac{\text{cm}^3}{\text{s}}\)

 

[karush]

Re: expanding cube

$$3\cdot 2^2 \cdot 6 = 72 \text { cm}^3\text{/ sec}$$

and

$$3\cdot 10^2 \cdot 6 = 1800 \text { cm}^3\text{/ sec}$$

 

[MarkFL]

Re: expanding cube

$$3\cdot 2^2 \cdot 6 = 72 \text { cm}^3\text{/ sec}$$

and

$$3\cdot 10^2 \cdot 6 = 1800 \text { cm}^3\text{/ sec}$$

Looks good. If you wish to be absolutely clear on an exam, I would write (in addition to showing your differentiation with respect to $t$ to obtain the formula):

a) \(\displaystyle \left. \frac{dV}{dt} \right|_{s=2\text{ cm}}=3\left(2\text{ cm} \right)^2\left(6\,\frac{\text{cm}}{\text{s}} \right)=72\,\frac{\text{cm}^3}{\text{s}}\)

b) \(\displaystyle \left. \frac{dV}{dt} \right|_{s=10\text{ cm}}=3\left(10\text{ cm} \right)^2\left(6\,\frac{\text{cm}}{\text{s}} \right)=1800\,\frac{\text{cm}^3}{\text{s}}\)

 

[karush]

Re: expanding cube

makes sense
I will post some more to see how close I am