# Multi variable problem.

• Jan 27th 2009, 07:48 AM
craigmain
Multi variable problem.
Hi,

I have just completed a set of problems using the chain rule, and have come across one I am very confused about.

Given K=1/2mvv
(v squared)
m = 10grams
v = 30cm/second
accel = 5cm/s/s

what rate is K changing. I don't know how to go about applying the chain rule to the function. I have tried differentiating with respect to m, then with respect to v and multiplying by vv (v squared) and m respectively.

The answer is 1500. I can't figure out the middle step. I am guessing that accel of 5m/s/s is not required. There is no accel in the formula.

Thanks
Regards
Craig.
• Jan 27th 2009, 07:56 AM
Mush
Quote:

Originally Posted by craigmain
Hi,

I have just completed a set of problems using the chain rule, and have come across one I am very confused about.

Given K=1/2mvv
(v squared)
m = 10grams
v = 30cm/second
accel = 5cm/s/s

what rate is K changing. I don't know how to go about applying the chain rule to the function. I have tried differentiating with respect to m, then with respect to v and multiplying by vv (v squared) and m respectively.

The answer is 1500. I can't figure out the middle step. I am guessing that accel of 5m/s/s is not required. There is no accel in the formula.

Thanks
Regards
Craig.

You are asked to find the rate of change of K with time, so you must differentiate with respect to time!

$\displaystyle \frac{dK}{dt} = \frac{d}{dt} (\frac{1}{2}mv^2 )$

Take the constants out of the integration:

$\displaystyle \frac{dK}{dt} = \frac{1}{2}m\frac{d}{dt} (v^2)$

You need to differentiate implicitly here!

$\displaystyle \frac{dK}{dt} = \frac{1}{2}m (2v \times \frac{dv}{dt})$

You should realise that dv/dt is the rate of change of velocity with time... also known as... acceleration!

Hence:

$\displaystyle \frac{dK}{dt} = \frac{1}{2}m (2v \times a)$

$\displaystyle \frac{dK}{dt} = mva$
• Jan 27th 2009, 09:19 PM
craigmain
Thanks, still confused, but not because of your explanation.
Hi,

I appreciate your answer very much. I still have some reading to do, as I still don't understand why m is considered a constant, and also why once differentiating v squared you still need to multiply by dv/dt.

I need to keep reading. I am not understanding the various rates of change and how they relate to one another in the same equation.
• Jan 28th 2009, 02:55 AM
Mush
Quote:

Originally Posted by craigmain
Hi,

I appreciate your answer very much. I still have some reading to do, as I still don't understand why m is considered a constant, and also why once differentiating v squared you still need to multiply by dv/dt.

I need to keep reading. I am not understanding the various rates of change and how they relate to one another in the same equation.

The mass is considered constant, because it is constant! The mass never changes. Ever. Throughout your entire problem, only 2 things ever change... K, and the velocity. Hence these are the only two variables which have derivatives in the problem.

You were differentiating with respect to time. That means that only variables which are functions of time can be differentiated. Only things that change with time can be differenatited wrt time.

The mass doesn't change with respect to time. If you draw a mass/time graph, it will be a vertical/horizontal line. Constant.

Velocity is a function of time. V = d/t.

K is a function of V, hence it is a function of time, since v is a function of time.

But nothing else is a function of time in this problem, they can be considered constants.
• Jan 28th 2009, 03:02 AM
craigmain
Thanks
Thanks,
I understand the problem now.