Question

Two variable are connected by the relation \(\Huge\frac {1}{p}+\frac {1}{q}=\frac {1}{k}\) where k is a constant.
Given that k=10 and p decrease at the rate of 2, find the rate of change of q when p=40.

Answer 1

Can you just solve it for q, take the derivative with respect to p, and then plug in your values?

Answer 2

That's my main problem lol

Answer 3

Okay well if
\[ \frac{1}{p} + \frac{1}{q} = \frac{1}{k} \]
then
\[q = \frac{pk}{p-k} \]
Can you differentiate that with respect to p?

Answer 4

\(\Huge\frac {k(k-p-\frac{p^3}{3})}{(k-p)^2}\)
That's my dq/dp lol is it possible to simplify it? ==

Answer 5

sorry it dq/dp=\(\Huge\frac {k(p-k-\frac{p^3}{3})}{(p-k)^2}\)
it shud be like this if i does not make any mistake

Answer 6

yup u have done mistake doing u/v rule in derivatives...
the numerator is completely incorrect.

Answer 7

oops sorry again mistakes
\(\Huge\frac {k(p-k+\frac {p^2}{2})}{(p-k)^2}\)

Answer 8

whats the derivative of numerator?
whats the derivative of denominator?
how can u ever get p^2/2 ?
are u integrating??

Answer 9

OMG i mixed up differentiate with integrate lol

Answer 10

now since u will only differentiate here, u will get your answer quite easily.

Answer 11

so is it
\(\Huge\frac{k(2p-k)}{(p-k)^2}\)?

Answer 12

isn't the numerator k(p-k)-pk ?

Answer 13

which simplifies to -k^2 only.?

Answer 14

numerator i've done is k(p-k)+pk
so i factorise
k(p-k+p)=
k(2p-k)

Answer 15

The final result should be
\[\frac{dq}{dp} = -\left( \frac{k}{p-k} \right)^2 \]

Answer 16

it should be - pk, not + pk.

Answer 17

O.o i try to check my error

Answer 18

ok i have it and trying to finish the question

Answer 19

To clarify, if you want
\[ \frac{dq}{dt} \]
then that will be
\[ \frac{dq}{dp} \cdot \frac{dp}{dt} \]

Answer 20

p decrease at the rate of 2
This statement means dp/dq=2? or dq/dp

Answer 21

That means
\[ \frac{dp}{dt} = -2 \]

Answer 22

okay i have done the question. The answer is 2/9. btw thanks both @Jemurray3 and @hartnn

Answer 23

Good job.