# Thread: Find corresponding percentage increase

1. ## Find corresponding percentage increase

If L cm is the length of a pendulum and t s the time of one complete swing, it is known that L = kt^2. If the length of the pendulum is increased by x%, x being small, find the corresponding percentage increase in time of swing.

My workings:
L = kt^2
dL/dt = 2kt
delta t = dt/dL x delta L
delta t = 1/2kt x x% = x/2kt%

The answer given is x/2%.

Where where where have I gone wrong?!

2. Hello darkvelvet
Originally Posted by darkvelvet
If L cm is the length of a pendulum and t s the time of one complete swing, it is known that L = kt^2. If the length of the pendulum is increased by x%, x being small, find the corresponding percentage increase in time of swing.

My workings:
L = kt^2
dL/dt = 2kt
delta t = dt/dL x delta L
delta t = 1/2kt x x% = x/2kt%

The answer given is x/2%.

Where where where have I gone wrong?!
If you've read my reply to your previous post, you'll probably see where you've gone wrong.

What you mean is:
$\dfrac{\delta L}{L}=x$%
I'm sure you can complete it now.

3. Alright, this is what I've got:

$L=kt^2$
$\dfrac{dt}{dL}\approx\dfrac{\delta t}{\delta L}$
$\dfrac{dt}{t} = \dfrac{1/2kt * \delta L}t$
$\dfrac{dt}{t} = \dfrac{1/2kt * \delta L}{\sqrt(L/K)}$

Am I right so far?

4. Hello darkvelvet

Do it in the same way as the previous question:

$L = kt^2$

$\Rightarrow \dfrac{dL}{dt}= 2kt$

$\Rightarrow \dfrac{dL}{L}=\dfrac{2kt\,dt}{L}$
$=\dfrac{2kt\,dt}{kt^2}$
Can you continue?

Just a quick question: does $\frac{\delta L}{L}$ = x% mean the same as $\frac{\delta L}{L}$ * 100% = x% ?