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The volume of a spherical balloon grows at a rate of $100\ cm^3/s\ $,what is the growing rate when the radius measures $50cm$.

I already know how to work this out, But I can't understand the problem 100%.

I was thinking about this problem and some doubts came up.

In the following example, the problem already give a related rate, that is, Volume over Time.

But then I thought, If the volume increases 100cm^3/s, The radius is also increasing, as the volume depends on the radius. But the radius will also grow only if the volume increases that is both measures are related.

As both rates are constants, if I change 50cm for 20 cm wouldn't the dr/dt be the same ?

I worked it out for both measures, and the results are not the same, I'm wondering why.

While the volume is increasing, the radius is increasing as well, so If the radius is increasing the dr/dt is changing ?

I thought only the radius could variate, not the dr/dt.

I'm very confused.

______edit__________

@RoddyMacPhee As the sphere depends on the $R^3$ and not only on the R, I can say that the variation of the radius of the sphere will never be proportional to the variation of the volume of the sphere, because the radius will never grow proportionally to the rate of variation of the volume, because the volume will increase 100 unidades of $\ cm^3 \ $ per second.

and the radius will always be increasing in a exponential way, that is impossible to determine a ratio between your variation per second.

So, the rate of increase of the radius when it measures 50cm, means how much it increased to get to 50cm, and not a default increase per time until it measures 50 cm.

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    $\begingroup$ The rates cannot both be constant. A constant rate of increase of radius will result in a non-constant rate of increase of volume and vice-versa. $\endgroup$
    – Shuri2060
    Commented Jul 30, 2017 at 21:04
  • $\begingroup$ @Shuri2060 Could you explain why ? thanks. $\endgroup$
    – Goun2
    Commented Jul 30, 2017 at 21:06
  • $\begingroup$ In addition, $\frac{dr}{dt}$ depends on $r$ in general (unless $\frac{dr}{dt}$ is constant). $\endgroup$
    – Shuri2060
    Commented Jul 30, 2017 at 21:07
  • $\begingroup$ For the above - let's say the rate of change of radius is constant. Then clearly the change of volume from $r=0$ to $r=1$ is smaller than from $r=1$ to $r=2$. $\endgroup$
    – Shuri2060
    Commented Jul 30, 2017 at 21:09
  • $\begingroup$ @hyx whats the volume formula for a sphere ... ? $\endgroup$
    – user451844
    Commented Jul 30, 2017 at 21:09

4 Answers 4

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From the formula for the volume of a sphere, we have $V=\dfrac{4}{3} \pi r^{3}$. This shows how the volume depends on the radius - when the volume is constant, the radius is constant. However, it does not show that when the rate of change of the volume is constant, the rate of change of the radius is constant.

If we differentiate with respect to t, we get the relationship between the rate of increase in the volume and the rate of increase of the radius (with respect to time) $\dfrac{dV}{dt}=4 \pi r^{2}\dfrac{dr}{dt}$. This shows that both cannot be constant as the radius is not constant.

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By differentiating the volume formula,

$$\begin{align*} V &= \frac23 (2\pi)r^3\\ \frac {dV}{dt} &= \frac23(2\pi)\cdot 3r^2\frac{dr}{dt} \end{align*}$$

If $\frac{dV}{dt}$ is constant and non-zero, $\frac{dr}{dt}$ is not constant and depends on the $r$ at that moment.

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hint

The volume is given by $$V (t)=\frac {4\pi}{3}R^3 (t) $$

thus

$$\frac {dV}{dt}=4\pi R^2\frac {dR}{dt} $$ with $$\frac {dV}{dt}=100\;\; cm^3/s $$ and $$R=50 \;\;cm $$ hence

$$\frac {dR}{dt}=\frac {1}{100\pi} \;\;cm/s $$

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You may be overthinking this. The volume of the sphere is

$$V=4\pi r^3/3\\ \&\\ V=\frac{dV}{dt}t $$

since $dV/dt$ is constant. Therefore,

$$r=\left(\frac{3}{4\pi}\frac{dV}{dt}t\right)^{1/3}\\ \frac{dr}{dt}=\left(\frac{3}{4\pi}\frac{dV}{dt}\right)^{1/3}\frac{1}{3t^{2/3}} $$

Here we notice that $dr/dt$ is infinite at $t=0$ when the growth begin.

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  • $\begingroup$ So, what does the growing rate of the radius when it measure 50cm means ?, If possible take a look at my edit. $\endgroup$
    – Goun2
    Commented Jul 30, 2017 at 23:30
  • $\begingroup$ @hjx First calculate the time to reac r=50, then get dr/ft. It gives dr/dt=1/100pi for r=50. $\endgroup$ Commented Jul 31, 2017 at 0:20
  • $\begingroup$ That wasn't the question, but ok. the dv/dt is 100 cm^3/s, it means the sphere will be increasing 100 units of cm^3 per second. The dr/dt is 1/100 pi cm/s, but for only when the radius is 50 cm, as this rate isn't exact for all possible measures of the radius, what does 1/100pi cm/s mean ? I think there's no way I can understand this, unfortunatelly math is over my head. $\endgroup$
    – Goun2
    Commented Jul 31, 2017 at 1:03
  • $\begingroup$ @hjx I don't understand your problem, I gave you all the equations to solve for r, dr/dt, and V for any dV/dt. It's time to accept an answer and move on. $\endgroup$ Commented Jul 31, 2017 at 2:44

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