Answer to Question #97348 in Calculus for Matthew Maffey

Question

The equation for the instantaneous voltage across a discharging capacitor is given by v=Voe^-t/t , where Vo is the initial voltage and t is the time constant of the circuit.

The equation for the instantaneous voltage across a discharging capacitor is given by v=Voe^-t/t , where Vo is the initial voltage and t is the time constant of the circuit.

a) Draw a graph of voltage against time for Vo =12v and t=2s, between t=0s and t=10s?

b) Calculate the gradient at t=2s and t=4s?

c) Differentiate v=12e^-t/2 and calculate the value of dv/dt at t=2s and t=4s?

d) Compare answers for part b and part c?

e) Calculate the second derivative of the instantaneous voltage (d^2 v/dt^2)?

Answer 1

Answer

b) 2.20727665; 0.81201170.

c) 2.20727665; 0.81201170.

d) They are equal.

e) 3e^-t/2.

Explanation

a) First of all, we must correct the equation for the instantaneous voltage across a discharging capacitor, taking into account the dimension analysis. This equation must be written in the form v=V₀e^-t/T.

We draw a graph of voltage against time for Vo =12v and T=2s, between t=0s and t=10s on the following Figure1.

Figure 1

b) Let us consider the problem of finding the gradient of a curve, and therefore its tangent. We can approximately calculate the gradient of a curve at the given point as the gradient of its chord, which goes from a near point. As we can see in the Figure 2, the gradient (slope) of chord is given by

BC/AC = (v(t) - v(a))/(t - a) = (v(a+h) - v(a))/h.

Figure 2

Calculating the limit of gradient of chord when h$\to$0 , we obtain the gradient of tangent by applying L’Hopital’s Rule rule

BC/AC =12(e^-a/2 e^-h/2-e^-a/2)/h= 12e^-a/2(e^-h/2 – 1)/h $\to$

$\to$ 12e^-a/2((e^-h/2)’ – (1)’)/(h)’ $\to$ -6e^-a/2 as h $\to 0.$

So at a = 2s the gradient of tangent is equal to

6e^-2/2 = 6/e = 2.20727665.

And at a = 4s the gradient of tangent is equal to

6e^-4/2 = 6/e^2 = 0.81201170

c) The derivative of v=12e^-t/2 is equal to

v’=12(e^-t/2)’ = 12e^-t/2(-t/2)’ = -6e^-t/2.

At t=2s the derivative is equal to

v’ = 6/e = 2.20727665.

And at t=4s

v’ = 6e^-t/2 = 6/e² = 0.81201170.

d) The comparison shows us that the gradient of tangent at the given point is equal to the function's derivative at this point.

e) The second derivative of the instantaneous voltage is equal to

(d^2 v/dt^2) = -6(e^-t/2)’ = -6e^-t/2 (-t/2)’=3e^-t/2.