Problem Overview
We calculate the rate of voltage change (d V d t \frac{dV}{dt} d t d V
V ( t ) = e − 2 t sin ( 3 t ) t 2 + 5 V(t) = \frac{e^{-2t} \sin(3t)}{\sqrt{t^2 + 5}} V ( t ) = t 2 + 5 e − 2 t sin ( 3 t )
evaluated at
t = 2 ms t = 2\text{ ms} t = 2 ms with
V ( 2 ) ≈ 0.0104 V V(2) \approx 0.0104\text{ V} V ( 2 ) ≈ 0.0104 V .
Step-by-Step Mathematical Solution
Step 1: Formally Apply Logarithmic Differentiation
The governing process is Logarithmic Differentiation :
d d t [ ln V ] = 1 V d V d t ⟹ d V d t = V ⋅ d d t [ ln ( V ( t ) ) ] \frac{d}{dt}[\ln V] = \frac{1}{V}\frac{dV}{dt} \implies \frac{dV}{dt} = V \cdot \frac{d}{dt}[\ln(V(t))] d t d [ ln V ] = V 1 d t d V ⟹ d t d V = V ⋅ d t d [ ln ( V ( t ))] (This rule is defined in Logarithmic Differentiation .)
Take the natural logarithm of both sides:
ln V = ln [ e − 2 t sin ( 3 t ) ( t 2 + 5 ) 1 / 2 ] \ln V = \ln\left[ \frac{e^{-2t} \sin(3t)}{(t^2 + 5)^{1/2}} \right] ln V = ln [ ( t 2 + 5 ) 1/2 e − 2 t sin ( 3 t ) ] Step 2: Expand Using Logarithmic Properties
Using the Product, Quotient, and Power properties:
ln V = ln ( e − 2 t ) + ln ( sin ( 3 t ) ) − ln ( t 2 + 5 ) 1 / 2 \ln V = \ln(e^{-2t}) + \ln(\sin(3t)) - \ln(t^2 + 5)^{1/2} ln V = ln ( e − 2 t ) + ln ( sin ( 3 t )) − ln ( t 2 + 5 ) 1/2 ln V = − 2 t + ln ( sin ( 3 t ) ) − 1 2 ln ( t 2 + 5 ) \ln V = -2t + \ln(\sin(3t)) - \frac{1}{2}\ln(t^2 + 5) ln V = − 2 t + ln ( sin ( 3 t )) − 2 1 ln ( t 2 + 5 ) Step 3: Differentiate Implicitly with Respect to Time t t t
1 V d V d t = d d t [ − 2 t ] + d d t [ ln ( sin ( 3 t ) ) ] − 1 2 d d t [ ln ( t 2 + 5 ) ] \frac{1}{V}\frac{dV}{dt} = \frac{d}{dt}[-2t] + \frac{d}{dt}[\ln(\sin(3t))] - \frac{1}{2}\frac{d}{dt}[\ln(t^2 + 5)] V 1 d t d V = d t d [ − 2 t ] + d t d [ ln ( sin ( 3 t ))] − 2 1 d t d [ ln ( t 2 + 5 )]
Apply the Chain Rule to each term:
1 V d V d t = − 2 + 3 cos ( 3 t ) sin ( 3 t ) − 1 2 ( 2 t t 2 + 5 ) \frac{1}{V}\frac{dV}{dt} = -2 + \frac{3\cos(3t)}{\sin(3t)} - \frac{1}{2}\left(\frac{2t}{t^2 + 5}\right) V 1 d t d V = − 2 + sin ( 3 t ) 3 cos ( 3 t ) − 2 1 ( t 2 + 5 2 t ) 1 V d V d t = − 2 + 3 cot ( 3 t ) − t t 2 + 5 \frac{1}{V}\frac{dV}{dt} = -2 + 3\cot(3t) - \frac{t}{t^2 + 5} V 1 d t d V = − 2 + 3 cot ( 3 t ) − t 2 + 5 t Step 4: Multiply by V V V d V d t \frac{dV}{dt} d t d V
d V d t = V ( t ) ⋅ [ − 2 + 3 cot ( 3 t ) − t t 2 + 5 ] \frac{dV}{dt} = V(t) \cdot \left[ -2 + 3\cot(3t) - \frac{t}{t^2 + 5} \right] d t d V = V ( t ) ⋅ [ − 2 + 3 cot ( 3 t ) − t 2 + 5 t ] Step 5: Evaluate at t = 2 ms t = 2\text{ ms} t = 2 ms V ( 2 ) ≈ 0.0104 V V(2) \approx 0.0104\text{ V} V ( 2 ) ≈ 0.0104 V
d V d t ∣ t = 2 = ( 0.0104 ) ⋅ [ − 2 + 3 cot ( 6 rad ) − 2 2 2 + 5 ] \left.\frac{dV}{dt}\right|_{t=2} = (0.0104) \cdot \left[ -2 + 3\cot(6\text{ rad}) - \frac{2}{2^2 + 5} \right] d t d V t = 2 = ( 0.0104 ) ⋅ [ − 2 + 3 cot ( 6 rad ) − 2 2 + 5 2 ]
In radian mode,
cot ( 6 ) ≈ − 3.4364 \cot(6) \approx -3.4364 cot ( 6 ) ≈ − 3.4364 :
d V d t ∣ t = 2 = ( 0.0104 ) ⋅ [ − 2 + 3 ( − 3.4364 ) − 2 9 ] \left.\frac{dV}{dt}\right|_{t=2} = (0.0104) \cdot \left[ -2 + 3(-3.4364) - \frac{2}{9} \right] d t d V t = 2 = ( 0.0104 ) ⋅ [ − 2 + 3 ( − 3.4364 ) − 9 2 ] d V d t ∣ t = 2 = ( 0.0104 ) ⋅ [ − 2 − 10.3092 − 0.2222 ] = ( 0.0104 ) ⋅ [ − 12.5314 ] ≈ − 0.1303 V/ms \left.\frac{dV}{dt}\right|_{t=2} = (0.0104) \cdot [ -2 - 10.3092 - 0.2222 ] = (0.0104) \cdot [-12.5314] \approx -0.1303\text{ V/ms} d t d V t = 2 = ( 0.0104 ) ⋅ [ − 2 − 10.3092 − 0.2222 ] = ( 0.0104 ) ⋅ [ − 12.5314 ] ≈ − 0.1303 V/ms Engineering Conclusion
At t = 2 ms t = 2\text{ ms} t = 2 ms − 0.1303 V/ms -0.1303\text{ V/ms} − 0.1303 V/ms