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# Calculus

## Velocity and Displacement

From, derivatives, we know that velocity, v, is the first derivative, and acceleration, a, is the second derivative. We can express this as

$a=\frac{dv}{dt}$

If we rearrange the equation to $$dv=a\,dt$$, we can integrate in respect to t

$v=\int a\,dt$

If the acceleration is constant, we have $v = at+c_1$

A similar equation can be set up to find the displacement, s, in terms of time.

$s=\int v\,dt$

Example: During the initial stage of launching a spacecraft vertically, the acceleration $$a\,(in\,m/s^2)$$ of the spacecraft is $$a=6t^2$$. Find the height $$s$$ of the spacecraft after $$6.0s$$ if $$s=12m$$ for $$t=0s$$ and $$v=16\,m/3$$ for $$t=2s$$.

Solution

First, we find the equation of the veloctiy.

\begin{align} v&=\int 6t^2\,dt=2t^3+c_1 \\ 16&=2(2)^3+c_1 \end{align}

We solve and find that $$c_1 =0$$, substitute $$c_1$$ at $$t=2s$$ back in and $$v=2t^3$$.

Next we find the equation for the displacement.

\begin{align} s&=\int 2t^3\,dt=\frac{1}{2}t^4+c_2 \\ 12&=\frac{1}{2}(0)^4+c_2 \end{align}

We solve and find that $$c_2 =12$$, substitute $$c_2$$ at $$6.0s$$ back in and $$s=\frac{1}{2}(6)^4+12=660m$$.

## Voltage Across a Capacitor

By definition, the current i in an electric circuit equals the time rate of change of the charge q (in coloumbs) that passes a given point in the circuit.

\{i=\frac{dq}{dt}\]

Thus,

$q=\int i\,dt$

The voltage $$V_C$$ across a capacitor with capacitance C is given by $$V_C=q/C$$. Combining the equations the voltage is given by

$V_C=\frac{1}{C} \int i\,dt$

The measurements $$V_C$$ is in volts, $$C$$ is in farads, $$i$$ in amperes, and $$t$$ in seconds.

Example: A certain capacitor has $$100 V$$ across it. At this instant, a current $$i=0.06t^{1/2}$$ is sent through the circuit. After $$0.25s$$, the voltage across the capacitor is $$140V$$. What is the capacitance?

Substituting $$i=0.06t^{1/2}$$ , we find that,

\begin{align} V_C &= \frac{1}{C}\int 0.06t^{\frac{1}{2}}dt \\ &= \frac{0.06}{C}t^{\frac{1}{2}}dt \\ &= \frac{0.04}{C}t^{\frac{3}{2}}+c_1 \end{align}

Now we solve for $$c_1$$ with $$V_C=100$$ at $$t=0$$ and find that $$c_1=100V$$. Thus,

$$V_C=\frac{0.04}{C}t^{\frac{3}{2}}+100$$

We also know that $$V_C=140V$$ at $$t=0.25s$$

\begin{align}140 &=\frac{0.04}{C}(0.25)^{\frac{3}{2}}+100 \\ 40&=\frac{0.04}{C}(0.125) \\ C&= 125\times 10^{-4} F=125\mu F\end{align}