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- Calculus
- Limits
- Continuity
- Definition of the Derivative - First Principles
- Basic Differentiation Rules
- More Differentiation Rules
- Implicit Differentiation
- Higher Order Derivatives
- Curve Sketching
- First Derivative Test
- Second Derivative Test
- Derivatives of Trigonometric Functions
- Derivatives of Inverse Trigonometric Functions
- Derivatives of Logarithmic Functions
- Derivatives of Exponential Functions
- Antiderivative
- Indefinite Integral
- Applications of the Indefinite Integral
- The Definite Integral
- Area Under a Curve - Riemann Sums
- Area Under a Curve - Integration
- Area between Two Curves
- Volumes of Revolution
- Logarithmic Integrals
- Exponential Integrals
- Trigonometric Integrals
- Trigonometric Integrals of Other Forms
- More Integration Methods

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 \(s=\frac{1}{2}(6)^4+12=660m\).

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}

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

- Last Updated: Aug 12, 2022 11:32 AM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717032
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