Question

How to solve this derivation http://s26.postimg.org/586cugd7d/Question.png

Answer 1

apply this rule
the diff. of
f/g=(f’ g - g’ f )/g^2

Answer 2

What are you stuck on Mr Hanshen?
Is it just a little confusing because it's a partial derivative?

Answer 3

\(\Large\bf\sf c_a\) and \(\Large\bf\sf m_a\) are constant?

Answer 4

yes, I'm confuse about the partial derivative because there is substition in that..
the ca and ma are constant, T is changing

Answer 5

\[\Large\bf\sf C_k\quad=\quad c_a m_a\frac{T_a-T_m}{T_m-T_k}\]

\[\large\bf\sf \frac{\partial C_k}{\partial T_m}\quad=\quad c_a m_a\frac{\color{royalblue}{\left(T_a-T_k\right)_{T_k}}(T_m-T_k)-(T_a-T_k)\color{royalblue}{\left(T_m-T_k\right)_{T_k}}}{(T_m-T_k)^2}\]

Answer 6

We set it up as quotient rule.
We need to take the derivative of these blue parts (partial with respect to T_m)

Answer 7

Ugh i made some bad typos in there :( lemme try to fix that...

Answer 8

Mmmm ok I think that looks better :O

Answer 9

So for the first blue part,
We treat T_a as a constant when we differentiate.\[\Large\bf\sf \frac{\partial }{\partial T_m}(T_a-T_m)\quad=\quad 0-1\]

Answer 10

\[\large\bf\sf \frac{\partial C_k}{\partial T_m}\quad=\quad c_a m_a\frac{\color{royalblue}{\left(T_a-T_m\right)_{T_m}}(T_m-T_k)-(T_a-T_m)\color{royalblue}{\left(T_m-T_k\right)_{T_m}}}{(T_m-T_k)^2}\]Bahhh I still missed one! lol.
Ok ok I think it's ok now

Answer 11

i see, how about everything is constant? what is the difference on the answer?

Answer 12

Hmm we wouldn't really want to do that.
\[\Large\bf\sf \frac{\partial C_k}{\partial T_m}\quad \text{Translates: How fast is }C_m\text{ changing?}\]If everything is being held constant, then nothing is changing.
The rate of change is zero.\[\Large\bf\sf \frac{\partial C_k}{\partial T_m}\quad=\quad 0\]

Answer 13

UGHH C_k, not C_m.
I can't keep these subscripts straight :( Too many!!

Answer 14

I see, how about this answer. \[-\frac{ ca.ma ((Tm-Tk)+(Ta-Tm)) }{ (Tm-Tk)^{2} }\]
is this answer correct? I'm confuse how my friend can make it into this.
do i miss something to make the derivation look like my friend, or is there anychance my friend is wrong?

Answer 15

Yes that looks correct.
But the problem can be simplified further than that.
You don't want to stop there.

Answer 16

ah i think i got it now. the blue one become 1 afther the derivation.
ow yeahhhh...

Answer 17

yes i want to input my data into that

Answer 18

Quotient rule setup,
\[\large\bf\sf \frac{\partial C_k}{\partial T_m}\quad=\quad c_a m_a\frac{\color{royalblue}{\left(T_a-T_m\right)_{T_m}}(T_m-T_k)-(T_a-T_m)\color{royalblue}{\left(T_m-T_k\right)_{T_m}}}{(T_m-T_k)^2}\]Taking our partials,\[\large\bf\sf \frac{\partial C_k}{\partial T_m}\quad=\quad c_a m_a\frac{\color{orangered}{\left(-1\right)}(T_m-T_k)-(T_a-T_m)\color{orangered}{\left(1\right)}}{(T_m-T_k)^2}\]

Answer 19

We have some uhhh, T_m's canceling out, yes?

Answer 20

what is cancelling out?

Answer 21

The first partial is giving you a -1.
When you distribute all the negative signs in the problem,
you end up with a numerator like this:\[\large\bf\sf \frac{\partial C_k}{\partial T_m}\quad=\quad c_a m_a\frac{-T_m+T_k-T_a+T_m}{(T_m-T_k)^2}\]

Answer 22

-Tm + Tm, they "cancel out", yes?
-Tm + Tm = 0

Answer 23

ah.. i see... that's better haha

Answer 24

I can continue my work now. Thanks alot zepdrix :)

Answer 25

thanks for your time.

Answer 26

No prob \c:/