Calculus Problem

Mathmasteriw

Junior Member
Joined
Oct 22, 2020
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85
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 ?
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I have seen that in many places.
dvdt(5)=dv(5)dt\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.
 
I have seen that [derivative notation] in many places …
Thanks, Halls. It prompted me to think of and ponder this:

dvdt5=0.164\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 [  dvdt(5)   \; \frac{dv}{dt}(5) \; ] to [  dv(5)dt   \; \frac{dv(5)}{dt} \; ] …
Agree!

?
 
How I look at it is that dvdt\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. dvdt=dvdt(t)\dfrac{dv}{dt}=\dfrac{dv}{dt}(t) and dvdt\dfrac{dv}{dt} evaluated at t= 5 is denoted as dvdt(5)\dfrac{dv}{dt}(5).

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