Use differentiation to find the rate of change

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fredyneedshelp

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Hi there, basically I've been on this for hours, watched many videos and do understand the basics however this just throws me.

So ive been given the charging voltage formula and need to differentiate it to find the rate of change of voltage at 6 ms.

Now I can plug in numbers to equations buts it differentiating the formula with chain rule and product rules am having trouble with.

Charge formula is v=V(1-e^-t/T) where T=CR and is the time constant

So ive worked out so far v(t)=V(1-e^-t/T)

So to multiply out of brackets v(t)=V-Ve^-t/T

The first V being a constant so v(t)=0-Ve^-t/T

Which leaves me with the -Ve^-t/T

Its the negatives that are throwing me before the Ve and also the divided negative exponent.

I have been told the answer finished is V/T . e^-t/T

I am not one to just write down an answer without fully understanding why it is this.

Any help right now would be very much appreciated.

Thank you
 
fredyneedshelp said:
Hi there, basically I've been on this for hours, watched many videos and do understand the basics however this just throws me.

So ive been given the charging voltage formula and need to differentiate it to find the rate of change of voltage at 6 ms.

Now I can plug in numbers to equations buts it differentiating the formula with chain rule and product rules am having trouble with.

Charge formula is v=V(1-e^-t/T) where T=CR and is the time constant

So ive worked out so far v(t)=V(1-e^-t/T)

So to multiply out of brackets v(t)=V-Ve^-t/T

The first V being a constant so v(t)=0-Ve^-t/T

Which leaves me with the -Ve^-t/T

Its the negatives that are throwing me before the Ve and also the divided negative exponent.

I have been told the answer finished is V/T . e^-t/T

I am not one to just write down an answer without fully understanding why it is this.

Any help right now would be very much appreciated.

Thank you
Click to expand...
You need to write:

v(t) = V * [1-e^(-t/T)]

\(\displaystyle \displaystyle {v(t) \ = \ V * [1-e^{-\frac{t}{T}}]}\)

\(\displaystyle \displaystyle {\dfrac{d}{dt}[v(t)] \ = \ V * [-\dfrac{1}{T}] * [-e^{-\dfrac{t}{T}}]] \ = \ V * [\dfrac{1}{T}] * [e^{-\dfrac{t}{T}}]]}\)
 
Hi there, basically I've been on this for hours, watched many videos and do understand the basics however this just throws me.

So ive been given the charging voltage formula and need to differentiate it to find the rate of change of voltage at 6 ms.

Now I can plug in numbers to equations buts it differentiating the formula with chain rule and product rules am having trouble with.

Charge formula is v=V(1-e^-t/T) where T=CR and is the time constant

So ive worked out so far v(t)=V(1-e^-t/T)

So to multiply out of brackets v(t)=V-Ve^-t/T

The first V being a constant so v(t)=0-Ve^-t/T
Click to expand...
No, the derivative of a constant is 0. You don't yet have the derivtive

Which leaves me with the -Ve^-t/T

Its the negatives that are throwing me before the Ve and also the divided negative exponent.
Click to expand...
The derivative of Af(g(t)), where A is a constant, f and g are functions is, by the chain rule, is Af'(g(t))g'(t).
Here, A= -V, \(\displaystyle f(g(t) e^{-t/T}\) so we can take \(\displaystyle f(t)= e^t\) and \(\displaystyle g(t)= -t/T\). Then \(\displaystyle f'(t)= e^t\) so \(\displaystyle f'(g(t))= f'(-t/T)= e^{-t/T}\) and \(\displaystyle g'(t)= -1/T\). \(\displaystyle Af'(g(t))g'(t)= -Ve^{-t/T}(-1/T)= (V/T)e^{-t/T}\).

I have been told the answer finished is V/T . e^-t/T

I am not one to just write down an answer without fully understanding why it is this.

Any help right now would be very much appreciated.

Thank you
Click to expand...
 
Just in case Subhotosh's very elegant explanation slips by you and since you mentioned the chain rule, you can use substitution of variables as an alternative approach.

[MATH]v = V(1 - e^{-(t/T)}),\ u= e^w \text { and } w = -\ \dfrac{t}{T} = t * \left (-\ \dfrac{1}{T} \right ) \implies[/MATH]
[MATH]v = V(1 - u) = V - Vu,\ \dfrac{du}{dw} = e^w = e^{(-(t/T)},\ \text { and } \dfrac{dw}{dt} = -\ \dfrac{1}{T}.[/MATH]
[MATH]\therefore \dfrac{dv}{dt} = 0 - u(0) - V \dfrac{du}{dt} = -\ V * \dfrac{du}{dw} * \dfrac{dw}{dt} =[/MATH]
[MATH]-\ V * e^{-(t/T)} * \left (-\ \dfrac{1}{T} \right ) = \dfrac{V e^{-(t/T)}}{T}.[/MATH]
Obviously this is equivalent to the answers in the two previous posts, but substitution of variables is a very mechanical way to proceed that helps to avoid errors.
 
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