Question

Help with numer 5 please. thank you​

Answer 1

Answer:

See Below.

Step-by-step explanation:

We are given that:

$\displaystyle I = I_0 e^{-kt}$

Where I₀ and k are constants.

And we want to prove that:

$\displaystyle \frac{dI}{dt}+kI=0$

From the original equation, take the derivative of both sides with respect to t. Hence:

$\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right]$

Differentiate. Since I₀ is a constant:

$\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)$

Using the chain rule:

$\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right) = -kI_0e^{-kt}$

We have:

$\displaystyle \frac{dI}{dt}+kI=0$

Substitute:

$\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0$

Distribute and simplify:

$\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0$

Hence proven.