Turn on thread page Beta
 You are Here: Home >< Maths

# Rate of change watch

1. A right circular cone has a constant volume. The height h and the base radius can vary. Find the rate at which h is changing with respect to r at the instant when h and r are equal.
Ok so volume of cone is Πr^2(h/3)
I believe the question asked for dh by dr when r=h
So if r=h , v= Π(r^3)/3
So dv/dr = Πr^2
Same thing for dv/dh= Π(h^2)
Using chain rule to get dh/dr= r^2/h^2

Im not getting the textbook answer
Textbook answer is -2
2. (Original post by Bilbao)
A right circular cone has a constant volume. The height h and the base radius can vary. Find the rate at which h is changing with respect to r at the instant when h and r are equal.
Ok so volume of cone is Πr^2(h/3)
I believe the question asked for dh by dr when r=h
So if r=h , v= Π(r^3)/3
So dv/dr = Πr^2
Same thing for dv/dh= Π(h^2)
Using chain rule to get dh/dr= r^2/h^2

Im not getting the textbook answer
Textbook answer is -2
That's because you're subbing in r=h right away without really thinking about it. The two variables change change at different rates, by making them equal before differentiation you assume they change at the same rate.

We have which is by the product rule + implicit differentiation.

We have the LHS = 0 because volume remains constant, hence at r=h we get ... ?

ALTERNATIVELY: We know V = const. so we can arrange the volume so that hence and since the volume is when we get the answer this way too.
3. (Original post by RDKGames)
That's because you're subbing in r=h right away without really thinking about it. The two variables change change at different rates, by making them equal before differentiation you assume they change at the same rate.
But aren't they changing at the same rate?
If r increases, h will have to decrease by the same amount for V to remain constant i.e get the same value for any change in r and h?
4. Alright. Thx
5. (Original post by Carlos Nim)
But aren't they changing at the same rate?
If r increases, h will have to decrease by the same amount for V to remain constant i.e get the same value for any change in r and h?
If over a second then for the volume to remain the same we must have over the same second ie radius doubles while the height goes down to 1/4 of its size. So no, the rates at which they change are not the same. Also, they can't be because one rate must be +ve and the other must be -ve!

Turn on thread page Beta

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 12, 2018
Today on TSR

### Predict your A-level results

How do you think you'll do?

### University open days

1. University of Bradford
Wed, 25 Jul '18
2. University of Buckingham
Psychology Taster Tutorial Undergraduate
Wed, 25 Jul '18
3. Bournemouth University
Clearing Campus Visit Undergraduate
Wed, 1 Aug '18
Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE