# Thread: rates of change - voltage over time

1. ## rates of change - voltage over time

I am having trouble getting my head around this question. please tell me if i am on the right track, and no idea how to go about part b.

the voltage across a resistor in a circuit is given by V=IR
if R= 0.010t^2 (ohms) and I= 4.12+.020t (amps) t = time in seconds

a) find the rate of change of the voltage with respect to time when t=2.5s

v= (4.12+.020t)(0.010t^2)

expanded v = 0.0412t^2 + 0.0002t^3

dv/dt = 0.0824t + 0.0006t^2
= 0.0824(2.5) + 0.0006(2.5)^2
= 0.20975 volts/second (210mv/sec)

b)at what time(s) will the voltage be a minimum?

????????????????????????? how do i turn this into an equation of a line? do i need to?

2. Ok, so I haven't checked your calculations for a), but your method is right

Right, part b).

You've got the equation for voltage, you worked it out to be v = 0.0412t^2 + 0.0002t^3. You want to find the minimum of this. So if you were to plot a graph for voltage, the minimum would be the at the lowest point, where the gradient is zero. So you've got an equation for the gradient, you worked out
dv/dt = 0.0824t + 0.0006t^2. You need to let this equal zero and solve for t. If you get more than one value out for t, work out which one is the minimum (the other will be a maximum).

Hope this helps.