# Math Help - rate confusion

1. ## rate confusion

the side of the cubes are increasing at the rate of 0.4cm per second. find:
1.the rate the volume increases
2.the rate the surface area is increasing when sides are 10cm

here's how i went about it.
v=s^3 where v is volume, s is length of the side
dv/dt = 3s^2 (ds/dt)

so from the question, the sides are increasing at a rate of 0.4cm per second.
from that you can get ds/dt i think. so how would i write that as ds/dt?
i'm guessing ds/dt = 1.04s as it increases by 4% persecond if the side was intially 1cm but i have no idea how to write ds/dt generally,

if i solve the first question then i'm pretty sure i can do part two.

thanks in advanced for any help

2. you are over complicating it ds/dt = 0.4cm/sec period.

3. Originally Posted by Calculus26
you are over complicating it ds/dt = 0.4cm/sec period.
what i don't understand is that when we multiply the rate in (0.4cm/sec). isn't it decreasing, not increasing by 0.4cm? since its a constant that is less than 1

so anyways tell me if this is right?
1. dv/dt= 3s^2 * 0.4
now what?

4. When you say $\frac{ds}{dt} = 0.4 \mbox{ cm/sec}$ it means that for every 1 second that passes, the side of the cube will increase by 0.4 cm. In other words, if the cube has a side of 5 cm at one moment, then exactly one second later, it will have a side of 5.4 cm. It is only decreasing if $\frac{ds}{dt} < 0$.

It can seem odd to take that number $0.4 \mbox{ cm/sec}$ and multiply it, but this is because are we just comparing the relationship between two numbers:

$\frac{ds}{dt}$ : how much the side increases per second

$\frac{dv}{dt}$ : how much the volume of the cube increases per second

Calculus tells us that there is a relationship between the two values above, which makes sense because clearly changing the side affects the volume, and vice versa.

That relationship is defined by this equation:

$\frac{dv}{dt} = 3s^2 \frac{ds}{dt}$

I'm not sure if this short explanation clears it up at all... let me know if you are still confused about it.

Moving on though, as far as solving the problem above, you must know at what value of $s$ we are solving this equation. The problem says at $s=10$, so we can plug in that information:

$\frac{dv}{dt} = 3s^2 \frac{ds}{dt}$

$\implies \frac{dv}{dt} = 3(10)^2 (0.4)$

Does this make sense?

5. i understand now thanks!
1.the rate the volume increases
-so in other words this question is inadequate to go further

ty

6. You could just say that the rate the volume increases is equal to $1.2s^2$, but you can't give a numerical answer without knowing what value of $s$ we are interested in.

7. Originally Posted by purebladeknight
the side of the cubes are increasing at the rate of 0.4cm per second. find:
1.the rate the volume increases
2.the rate the surface area is increasing when sides are 10cm

here's how i went about it.
v=s^3 where v is volume, s is length of the side
dv/dt = 3s^2 (ds/dt)

so from the question, the sides are increasing at a rate of 0.4cm per second.
from that you can get ds/dt i think. so how would i write that as ds/dt?
i'm guessing ds/dt = 1.04s as it increases by 4% persecond if the side was intially 1cm but i have no idea how to write ds/dt generally,

if i solve the first question then i'm pretty sure i can do part two.

thanks in advanced for any help

think it through simply.
Imagine a cube of any arbitrary size.
The sides increase by 0.4cm per second.
You cannot tell the exact increase in volume,
as the cube could be any size, but you have expressed it's volume
increase "in terms of" the side lengths and the increase in side length.
This is the answer to part 1.

$\frac{dV}{dt}=3s^2(0.4)cm^3\ per\ second.$

For part 2, you are asked to calculate the rate of change of surface area
given that s=10cm.

Surface area of a regular cube is $6s^2$ as it has 6 sides.

$\frac{d}{dt}A=\frac{d}{ds}6s^2\frac{ds}{dt}=12s(0. 4)cm^2\ per\ sec.$

Now use the given value for s.