Calculus Problem

[Mathmasteriw]

Hi guys and girls,
Can anyone double check my work here?
I belive I am correct here.
Differentiate the funcion with respect to ‘t’ and have shown the ‘rate of change’ when t is 5
Thanks for your time

 

[skeeter]

looks fine

 

[Otis]

\(\displaystyle \frac{dv}{dt}(5)\) …

Hi Mathmasteriw. That's an interesting form of function notation that I've never seen before. I'm curious. Where did you see it?

 

[HallsofIvy]

I have seen that in many places.
\(\displaystyle \frac{dv}{dt}(5)= \frac{dv(5)}{dt}\).
meaning the derivative of v with respect to t, evaluated at t=5
Some people would prefer the first to the second on the grounds that the second could be interpreted as the derivative of the constant v(5) which is, of course, 0.

 

[Jomo]

Personally I think the answer is (dv/dt)(5) = -2e^(-2.5) or -2/e^2.5. -.164 is an approximation and not the exact answer.

 

[Otis]

… -0.164 is an approximation and not the exact answer.

Good point, Jomo, but we don't know for sure what the given instructions may have said.

 

[Otis]

I have seen that [derivative notation] in many places …

Thanks, Halls. It prompted me to think of and ponder this:

\[\frac{\text{dv}}{\text{dt}} \bigg\rvert_{5} = -0.164\]

I've never seen that done, either, but it makes sense.

… Some people would prefer [\( \; \frac{dv}{dt}(5) \; \)] to [\( \; \frac{dv(5)}{dt} \; \)] …

Agree!

 

[Jomo]

How I look at it is that \(\displaystyle \dfrac{dv}{dt}\) is a function of t. Just because the name of the function is a bit strange looking does not change its notion. \(\displaystyle \dfrac{dv}{dt}=\dfrac{dv}{dt}(t)\) and \(\displaystyle \dfrac{dv}{dt}\) evaluated at t= 5 is denoted as \(\displaystyle \dfrac{dv}{dt}(5)\).

I have used this notation frequently. I have also seen \(\displaystyle \dfrac{dv}{dt}(5)\) written as \(\displaystyle \dfrac{dv}{dt}|_{t=5}\)

 

[Mathmasteriw]

Hi Mathmasteriw. That's an interesting form of function notation that I've never seen before. I'm curious. Where did you see it?

I owe a lot to what I have learned to this page

 

[topsquark]

\(\displaystyle \dfrac{dv}{dt}|_{t=5}\)

If memory serves me I've seen this in many forms in Math texts but the above quote is the only way I've seen Physicists write it.

-Dan