Engineering The voltage(V) volts in a circuit

[BonBon101]

Homework Statement

The voltage(V) volts in a circuit at time(t) is given by

V(t)=18(1 - e -4t/9)

Find the expression for dV/dt.

Hence evaluate dV/dt at

(a)t=0 and (b)t=5 seconds

Homework Equations

Just wondering if this is right and if not where did i go wrong

The Attempt at a Solution

dV/dt=18(-4/9)(1-e-4/9t)

= -9(1-e-4/9t)

(a) t=0 -8(1-e-4/9(0))

=0mA

(b)t=5 -8(1-e-4/9(5))

=-7.13mA

 

[Feldoh]

\frac{d}{dt}V(t) = \frac{d}{dt}18 - \frac{d}{dt}18e^{-4t/9} what's the derivative?

 

[BonBon101]

not to sure.Im sorry but I am not that good at this hoping to get better!

 

[Feldoh]

Well

\frac{d}{dt}V(t) = \frac{d}{dt}18 - \frac{d}{dt}18e^{-4t/9}

Whats the derivative of 18 with respect to t?

What about 18exp(-4t/9) with respect to t?

I separated the parts of the equation using the fact that \frac{d}{dt}[v(t) + w(t)] = \frac{d}{dt}[v(t)] + \frac{d}{dt}[w(t)] you can generally remember this rule as "the derivative of the sum is the sum of the derivatives".

Does this make sense?