Calculus problem differentiation.

[Lejas90210]

<Moderator's note: Member has been warned to show some effort before an answer can be given.>

Hello all.
This is my first post in this forum, I am asking for your understanding. I have a problem with the calculus task and I stuck in a dead endso I managed to find a solution on the internet. I am not sure whether it is done correctly. My main questions are:
1.) Why 9.5 * e ^ -1 = 3.495?
2.) Are the values in the table correct? What is the dependence between those values.
3.) If in response to t = 10, -> V (t) = 3.495 and in the table for t = 10 there is value: 60.05 V?

 

[Charles Link]

## e^{-1}=\frac{1}{2.7828} \approx.36788 ##.
## V'(10)=3.495 ##. (##Note: V'(t)=\frac{dV(t)}{dt} ##). That's the slope of the curve of ## V(t) ## vs. ## t ## at ## t=10 ## if you draw a tangent line. Note: The graph needs to have the y-increment the same as the x-increment to readily see this. (With your increments of 10 and 5, it will appear to have a slope of 3.495/2 at ## t=10 ##).

 

[Lejas90210]

My Problem (Task) require use different rule of differentiation that's what I have till now:

 

[Charles Link]

You need to realize that ## \frac{dV(t)}{dt} ## is called ## V'(t) ## and is completely different from ## V(t) ##.
If ## V(t) ## were distance, ## V'(t) ## would be the velocity.
## V(t) \neq V'(t) ##. They are two separate functions.
Again ## V'(t)=\frac{d V(t)}{dt} ##.
You can let ## V=y ## , and ## t=x ##, but you should specify this if you chose to take ##V'=y'= \frac{dy}{dx} ##. You then substitute ## V ## (or ## V' ##) and ## t ## back in, to get ## V'(t) ##.
It is good to be systematic rather than have hand-waving in your steps.