Rate of change problem with and without Leibniz notation

Question

I'm trying to learn Leibniz notation, so I'm trying to understand how this problem could be solved with and without Leibniz notation so I might get a connection with what I know. I only know Lagrange prime mark ($'$) notation for differentiation, but I'm not even sure how to solve this problem with Lagrange notation. Is that simply impossible or should I just focus on learning Leibniz notation?

Air is being pumped into a spherical balloon. The volume of the balloon is increasing at a rate of $20 \dfrac{\mathrm{cm}^3}{s}$ when the radius is $30 \;\mathrm{cm}$. How fast is the radius increasing at that time?

Adriano · Accepted Answer · 2014-12-04 17:36:37Z

Either notation is fine, although I would say more people tend to use Leibniz notation when dealing with Related Rates problems. The caveat for the other notation is to keep in mind what variable you are differentiating with respect to (usually, time).

We're given that $V' = \frac{dV}{dt} = 20 ~\mathrm{cm}^3 / \mathrm s$. We want to find $r'$ (or $\frac{dr}{dt}$) when $r = 30 ~\mathrm{cm}$. Now the volume $V$ of a sphere of radius $r$ is given by: $$ V = \frac{4\pi}{3}r^3 $$ Differentiating implicitly (with respect to $t$) yields: $$ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} $$ or alternatively: $$ V' = 4\pi r^2 \cdot r' $$ I'm sure you can take it from here.

Do I substitute $V'$ or $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ with $20\;\mathrm{cm}^3/\mathrm{s}$ and $r$ with $30\;\mathrm{cm}$? I'm still learning this so I just want to be sure, thanks in advance. — Frank Vel
– Frank Vel, Commented Dec 4, 2014 at 17:48

Frank Vel · Accepted Answer · 2014-12-04 18:37:36Z

\begin{array}{|crcl|rcl|}\hline &\text{Leibniz notation}&&&\text{Lagrange notation}\\\hline &V &=& \dfrac{2\tau r^3}{3}& V &=& \dfrac{2\tau r^3}{3}\\[1em] &\dfrac{\mathrm{d}V}{\mathrm{d}t} &=& 2\tau r^2\dfrac{\mathrm{d}r}{\mathrm{d}t}& V' &=& 2\tau r^2r'\\[1em] &\dfrac{1}{2\tau r^2}\dfrac{\mathrm{d}V}{\mathrm{d}t} &=& \dfrac{\mathrm{d}r}{\mathrm{d}t}& \dfrac{1}{2\tau r^2}V' &=& r'\\[1em]\hline r = 30\;\mathrm{cm}\Rightarrow&\dfrac{\mathrm{d}V}{\mathrm{d}t} &=& \dfrac{20\;\mathrm{cm}^3}{\mathrm{s}} & V' &=& \dfrac{20\;\mathrm{cm}^3}{\mathrm{s}}\\[1em]\hline &\dfrac{1}{2\tau (30\;\mathrm{cm})^2}\dfrac{20\;\mathrm{cm}^3}{\mathrm{s}} &=& \dfrac{\mathrm{d}r}{\mathrm{d}t}& \dfrac{1}{2\tau (30\;\mathrm{cm})^2}\dfrac{20\;\mathrm{cm}^3}{\mathrm{s}} &=& r'\\[1em] &\dfrac{1}{90\tau}\dfrac{\mathrm{cm}}{\mathrm{s}} &=& \dfrac{\mathrm{d}r}{\mathrm{d}t}& \dfrac{1}{90\tau}\dfrac{\mathrm{cm}}{\mathrm{s}} &=& r'\\\hline \end{array}

Stack Exchange Network

Rate of change problem with and without Leibniz notation

2 Answers 2

You must log in to answer this question.

Hot Network Questions