Question

the current is given by i=e^(-t * R/L) where R is the resistance in ohms, L is the inductance in henrys and t is the time in seconds. Calculate the rate of change of current, if the resistance decreases at 2Ohms, the inductance increases at 0.2 H/s at the instance when t=0.01s, L=0.5H and R=30Ohms

Answer 1

$I(t)$ is a composition of several functions, so we will find it's derivative (rate of change = derivative) by using the chain rule:

$\frac{d}{dt}I(t) = \frac{d}{dt}(e^{-t\cdot \frac{R}{L}})=e^{-t\cdot\frac{R}{L}} \cdot\frac{d}{dt}(-t\frac{R}{L})$ , as the derivative of an exponential is an exponential itself.

Now let's find the derivative of an expression in the brackets by using a derivative of a quotient:

$\frac{d}{dt}I(t)=e^{-t\cdot\frac{R}{L}} (- \frac{(Rt)'L-L'(Rt)}{L^2}) = e^{-t\cdot \frac{R}{L}} \cdot \frac{L'Rt-LR-LR't}{L^2}$

Now let's insert the given values : $R=30, L=0.5, t=0.01, L'=0.2, R'=-2$ (minus sign as R is decreasing):

$I'(t) = -e^{-0.6}\cdot\frac{0.06-15+0.01}{0.25} \approx -32.78 A/s$