Find the rate of increase of the radius of the pool

[chwala]

Homework Statement

see attached.

Relevant Equations

rate of change

My approach is as follows;

$\dfrac{dr}{dt}=\dfrac{dr}{da}⋅\dfrac{da}{dt}$

$\dfrac{dr}{da}=\dfrac{1}{2πr}$

$\dfrac{dr}{dt}=\dfrac{1}{2πr}⋅15$

$\dfrac{dr}{dt}=\dfrac{7.5}{πr}$

and noting that $r= \sqrt{\dfrac {50}{π}}$

$\dfrac{dr}{dt}=\dfrac{7.5}{12.533}=0.598$ cm/s

*kindly note that i do not have the solutions ..

 

[Orodruin]

Use units, please.

 

[Delta2]

Seems fine to me.

 

[Orodruin]

Seems fine to me.

That's 3 minutes after my post.

Homework Statement:: see attached.
Relevant Equations:: rate of change

View attachment 301874

My approach is as follows;

$\dfrac{dr}{dt}=\dfrac{dr}{da}⋅\dfrac{da}{dt}$

$\dfrac{dr}{da}=\dfrac{1}{2πr}$

$\dfrac{dr}{dt}=\dfrac{1}{2πr}⋅15$

$\dfrac{dr}{dt}=\dfrac{7.5}{πr}$

and noting that $r= \sqrt{\dfrac {50}{π}}$

$\dfrac{dr}{dt}=\dfrac{7.5}{12.533}=0.598$ cm/s

*kindly note that i do not have the solutions ..

I mean, use units throughout your calculation. Do not just slap them on at the end and assume that they are correct. For example, this

$\dfrac{dr}{dt}=\dfrac{1}{2πr}⋅15$

should read
$$
\frac{dr}{dt} = \frac{1}{2\pi r} (15\ {\rm cm}^2/{\rm s})
$$

 

[chwala]

Use units, please.

@Delta2 I had not included the units initially...I just amended. That's why this comment is valid.

@Orodruin I think the units at the end should be good enough...I don't think it's a standard math practise for students to show or rather indicate units at each and every step of their working.

 

[Delta2]

ok ok, but still think @Orodruin is a bit sort of pedantic but yeah if we want to be absolutely correct...

 

[Orodruin]

ok ok, but still think @Orodruin is a bit sort of pedantic but yeah if we want to be absolutely correct...

Answering with a number when the question asks for a dimensionful quantity is just not correct. Sure, you may assume that units match up but if 18 years of teaching has taught me anything, it is that you cannot assume that students get their units right. Therefore, I think it is good practice to be pedantic about following units throughout any computation. The next one may not be as obvious.

To figure the inches you’ve run,
Or to find the slug mass of the sun,
Forget your aversion
To unit conversion.
Just multiply (wisely!) by 1.
(https://www.physics.harvard.edu/undergrad/limericks)

 

[Mark44]

I mean, use units throughout your calculation. Do not just slap them on at the end and assume that they are correct.

Keeping units throughout a mathematical calculation is typically not done in math textbooks. For example, here is from a problem in "Calculus Concepts and Contexts," by James Stewart. The problem is about pumping air into a balloon at a certain rate, with a goal of finding the rate of change of the balloon's radius with respect to time.

Stewart includes units at the beginning:

Given: $\frac{dV}{dt} = 100 \frac{\text{cm}^2}{\text{sec}}$
Unknown: $\frac{dr}{dt}$ when r = 25 cm

None of the subsequent calculations include units. The final calculation line is $$\frac{dr}{dt} = \frac{1}{4\pi(25)^2} = \frac{1}{25\pi}$$
In his summary he includes the units of cm/sec.

A much older textbook I have, "Calculus and Analytic Geometry, 2nd Ed," by Abraham Schwartz (c. 1967), takes a similar tack in the three examples shown for related rates. There are no units shown in any of the calculations, and only the final summary of the answer includes units.

 

[Orodruin]

Keeping units throughout a mathematical calculation is typically not done in math textbooks. For example, here is from a problem in "Calculus Concepts and Contexts," by James Stewart. The problem is about pumping air into a balloon at a certain rate, with a goal of finding the rate of change of the balloon's radius with respect to time.

Stewart includes units at the beginning:

None of the subsequent calculations include units. The final calculation line is $$\frac{dr}{dt} = \frac{1}{4\pi(25)^2} = \frac{1}{25\pi}$$
In his summary he includes the units of cm/sec.

A much older textbook I have, "Calculus and Analytic Geometry, 2nd Ed," by Abraham Schwartz (c. 1967), takes a similar tack in the three examples shown for related rates. There are no units shown in any of the calculations, and only the final summary of the answer includes units.

That does not mean it is good practice …

The best practice imo is to use symbols to represent one’s quantities and only insert numbers at the very end.

 

[Mark44]

That does not mean it is good practice …

True enough. I was just presenting a couple of examples of what is very common in calculus textbooks. For the original problem in this thread, in which the work starts with the equation $\frac{dr }{dt } = \frac{dr }{dA}\frac{dA }{dt }$, one could easily confirm that the units on the left side are cm/sec, and the units on the right side are cm/cm^2 * cm^2/sec = cm/sec.

For what it's worth, a downside to including units at every step is that it clutters up the work, potentially making it more difficult to spot an incorrect step.

The best practice imo is to use symbols to represent one’s quantities and only insert numbers at the very end.

I agree. Both examples I cited following this practice, and when I was teaching calculus I always did likewise and urged my students to do so, as well.

 

[Orodruin]

True enough. I was just presenting a couple of examples of what is very common in calculus textbooks. For the original problem in this thread, in which the work starts with the equation $\frac{dr }{dt } = \frac{dr }{dA}\frac{dA }{dt }$, one could easily confirm that the units on the left side are cm/sec, and the units on the right side are cm/cm^2 * cm^2/sec = cm/sec.

That equation does not have a dimensional problem because the contained quantities have the correct physical dimension. Note that I say "physical dimension" and not "units" because what units you use to describe those physical dimensions are irrelevant to the validity. The left-hand side has dimension L/T and the right-hand side (L/L^2)(L^2/T) = L/T so it is dimensionally consistent.

A lot of times there will be some confusion regarding the difference between the physical dimensions and the units used to measure them.

For what it's worth, a downside to including units at every step is that it clutters up the work, potentially making it more difficult to spot an incorrect step.

This is why using defined dimensional quantities is better (such as Newton's constant $G$ or the Fermi constant $G_F$). When a numerical dimensionful quantity appears though, it is pretty standard to see it represented with units. Coulomb's law is
$$
F = k \frac{q_1 q_2}{r^2} = (8.988\cdot 10^9\ {\rm N m}^2/{\rm C}^2) \frac{q_1 q_2}{r^2},
$$
not
$$
(8.988\cdot 10^9) \frac{q_1 q_2}{r^2},
$$
which does not make sense dimensionally. It would only make sense if you add a describing sentence specifying "where the charges are measured in Coulomb, distance in meters, and force in Newtons" which is arguably even worse in terms of clutter.

 

[Prof B]

@Orodruin I think the units at the end should be good enough...I don't think it's a standard math practise for students to show or rather indicate units at each and every step of their working.

This is a physics forum. I learned to use units from a physics teacher. My math is better because of it.