Answer to Question #99177 in Calculus for M

Question

**MUST SHOW GRAPHICAL METHOD FOR CALCULATING GRADIENT**

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..

A) draw a graph that includes a graphical method for calculating gradient 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? (must be done on a numbered graph)

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

D) compare results for part b and c?

E) calculate the second derivative of the instantaneous voltage (d^2 v / dt^2)?

Answer 1

A)

We drew a tangent at $t=2$ and found the slope by finding x intercept and y intercept of the tangent

B)From the graph,

At $t=2s$

$x -intercept=2.6$

$y-intercept=3.2$

$Gradient = m= -tan(\theta)= -3.2/2.6=-1.23$

At $t=4s$

$x-intercept=4.7$

$y-intercept= 0.3$

$Gradient = m= -tan(\theta)= -0.3/4.7=-0.06$

C)$V=12e^{-t}/t$

$dV/dt=-12e^{-t}(1+t)/t^2$

At $t=2s$

$dV/dt=-1.23$

At $t=4s$

$dV/dt=-0.07$

D)$(m_2)_c>(m_2)_b$

$(m_4)_c>(m_4)_b$

D)$dV/dt=-12e^{-t}(1+t)/t^2$

$d^2V/dt^2=V_oe^{-t}(t^2+2t+2)/t^3$