1. ## Differential Analysis

For some reason, I'm not getting this....

The volume of an object is given by $\displaystyle V=[pi(4.5(t)+1)^2(t)]$ If t changes from 4 to 5:

Find dV and delta(V)

Heres how I am solving it:
Vprime= pi(4.5t+1)^2 +pi*t(2)(4.5t+1)(4.5)
dV= [Vprime(t)]*[dt].. dt=1 and t=4 so:

[pi(4.5(4)+1)^2 +pi*(4)(2)(4.5(4)+1)(4.5)][1]= 3282.96

deltaV=f(t+delta(t))-f(t) so:
deltaV=f(4+1)-f(4)=
[pi(4.5(5)+1)^2(5)]-[pi(4.5(4)+1)^2(4)]=4138.26

dV approximates delta(V) so the numbers should be very close but as you can see, they're not.

Can anyone help?

2. Originally Posted by Matt85

If t changes from 4 to 5

Vprime= pi(4.5t+1)^2 +pi*t(2)(4.5t+1)(4.5)
dV= [Vprime(t)]*[dt].. dt=1 and t=4 so:

[pi(4.5(4)+1)^2 +pi*(4)(2)(4.5(4)+1)(4.5)][1]= 3282.96

deltaV=f(t+delta(t))-f(t) so:
deltaV=f(4+1)-f(4)=
[pi(4.5(5)+1)^2(5)]-[pi(4.5(4)+1)^2(4)]=4138.26

dV approximates delta(V) so the numbers should be very close but as you can see, they're not.
The dV is correct. The delta(V) is correct.

The size of the change of V works with the size of the change of t. t increased by 25%, so of course V increased by a lot. If t went from 4 to 4.001, then you would see a more obvious connection.

EDIT: Thanks for the explanation

3. Originally Posted by Turiski
The derivative is correct. The delta(V) is correct

You assumed that dV was intended to be evaluated at t=4. You didn't seem to have any motivation for that, but perhaps there was some. In general though dV shouldn't be evaluated numerically.

Also, the size of the change of V works with the size of the change of t. t increased by 25%, so of course they are not particularly close. If t went from 4 to 4.001, then you would see a more obvious connection.
4 was the starting point, and the change in t (5-4) is 1. So, I was evaluating the change in the tangent line, dV, against the change in the actual equation, delta(V). Throughout this section, we've been comparing values like this.

Ahh, I get what you're saying about the change being more significant than from 4 to 4.001. And, it sounds like you came up with the same thing I did.