In order to use the Second Shift Theorem, the function needs to be entirely of the form $\displaystyle f\left( t - 1 \right) $. To do this let $\displaystyle v = t - 1 \implies t = v + 1 $, then

$\displaystyle \begin{align*}

\mathrm{e}^{-2\,t} &= \mathrm{e}^{-2 \, \left( v + 1 \right) } \\

&= \mathrm{e}^{-2\,v - 2 } \\

&= \mathrm{e}^{-2\,\left( t - 1 \right) - 2 } \\

&=...

Aidan's question via email about Fourier Transforms (2)]]>

In order to factorise this quadratic, we will need to recognise that $\displaystyle \mathrm{i}^2 = -1 $, so we can rewrite this as

$\displaystyle \begin{align*} \frac{1}{6 + 5\,\mathrm{i}\,\omega - \omega ^2 } &= \frac{1}{\mathrm{i}^2\,\omega ^2 + 5\,\mathrm{i}\,\omega + 6} \\ &= \frac{1}{ \left( \mathrm{i}\,\omega \right) ^2 + 5\,\mathrm{i}\,\omega + 6 } \\

&=...

Aidan's question via email about Fourier Transforms]]>

I'm assuming the hypothesis test is

$\displaystyle H_0 : \mu = 13 \quad \quad H_a : \mu < 13 $

We are given $\displaystyle \mu = 13, \quad \sigma = 2.47, \quad \bar{x} = 12.86 , \quad n = 20 $.

The test statistic is

$\displaystyle \begin{align*} z &= \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \\

&= \frac{12.86 - 13}{\frac{2.47}{\sqrt{20}}} \\

&\approx -0.253\,481 \end{align*} $

Thus the p value is

$\displaystyle \begin{align*} p &=...

Chloe's question via email about a p-value]]>

First we need to write the DE as a system of first order DEs.

Let $\displaystyle u = y $ and $\displaystyle v = y' $. Then

$\displaystyle \begin{align*} x^2\,y'' - 5\,x\,v + 7\,u &= 2\,x^3\ln{\left( x \right) }\\

x^2\,y'' &= 5\,x\,v - 7\,u + 2\,x^3\ln{ \left( x \right) } \\

y'' &= \frac{5\,x\,v - 7\,u + 2\,x^3\ln{\left( x \right) } }{x^2} \end{align*} $

So the system is

$\displaystyle \begin{align*} u' &= v , \quad \quad \quad \quad \quad...

Abdullah's question via email about Runge Kutta scheme]]>

We would need to recognise that the integral in the equation is a convolution integral, which has Laplace Transform: $\displaystyle \mathcal{L}\,\left\{ \int_0^t{ f\left( u \right) \,g\left( t - u \right) \,\mathrm{d}u } \right\} = F\left( s \right) \,G\left( s \right) $.

In this case, $\displaystyle g\left( t - u \right) = \sin{ \left[ 4\left( t - u \right) \right] } \implies g\left( t \right) = \sin{ \left( 4\,t \right) } \implies G\left( s \right)...

Abdullah's question via email about solving an integral equation using a Laplace Transform]]>

Taking the Laplace Transform of the equation gives

$\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 7 \left[ s\,Y\left( s \right) - y\left( 0\right) \right] + 6\,Y\left( s \right) &=...

Oscar's question via email about solving a DE using Laplace Transforms]]>

Since this is of the form $\displaystyle \frac{f\left( t \right)}{t} $ we should use $\displaystyle \mathcal{L}\,\left\{ \frac{f\left( t \right) }{t} \right\} = \int_s^{\infty}{F\left( u \right) \,\mathrm{d}u } $.

Here $\displaystyle f\left( t \right) = \cosh{\left( 4\,t \right) } - 1 $ and so

$\displaystyle F\left( s \right) = \frac{s}{s^2 - 16} - \frac{1}{s} $...

Seth's question via email about a Laplace Transform]]>

To apply this Runge-Kutta scheme, we will need to write our second order DE as a system of first order DEs.

Let $\displaystyle u = y $ and $\displaystyle v = y' $, then we have

$\displaystyle \begin{align*} y'' + 4\,v - 7\,u^2 &= 0.2 \\

y'' &= 0.2 - 4\,v + 7\,u^2 \end{align*} $

So our system of first order DEs is:

$\displaystyle \begin{align*} u' &= v , \quad \quad \quad \quad \quad \quad \quad \, u\left( 0 \right) = 3 \\

v' &= 0.2 - 4\,v + 7\,u^2...

Mahesh's question via email about a Runge-Kutta scheme]]>

The Bisection Method solves equations of the form $\displaystyle f\left( x \right) = 0 $ so we must write the equation as $\displaystyle 11\cos{ \left( x \right) } - 1 + 2\,\mathrm{e}^{-x/10} = 0 $. We can...

Lachlan's question via email about the Bisection Method]]>

The Secant Method is a numerical scheme to solve equations of the form $\displaystyle f\left( x \right) = 0 $, so we must rewrite the equation as $\displaystyle 0 = \frac{1}{2}\,x^2 - 10 - \sin{ \left( 1.8\,x \right) } $.

Thus...

Hayldiburasomas' question via email about Secant Method]]>

Newton's Method solves an equation of the form $\displaystyle f\left( x \right) = 0 $, so we need to rewrite the equation as

$\displaystyle \mathrm{e}^{1.2\,x} - 1.5 - 2.5\cos^2{\left( x \right) } = 0 $

Thus $\displaystyle f\left( x \right) =...

Alexander's question via email about Newton's Method]]>

The Bisection Method is used to solve equations of the form $\displaystyle f\left( x \right) = 0 $, so we need to rewrite the equation as $\displaystyle 8\cos{ \left( x \right) } - \mathrm{e}^{-x/7} = 0 $. Thus $\displaystyle f\left( x \right) =...

Lachlan's question via email about the Bisection Method]]>

You first have to write this DE as a system of first order equations.

Note, since $\displaystyle t$ does not appear in the original DE, that means that the system will be autonomous if kept in terms of $\displaystyle t$.

Let $\displaystyle y = u$ and $\displaystyle y' = v$, then

$\displaystyle \begin{align*}

y'' + 4\left( y' \right) ^2 - 7\,y &= 0.1 \\

y'' + 4\,v^2 - 7\,u &= 0.1 \\

y'' &= 7\,u - 4\,v^2 + 0.1

\end{align*}$

Thus the system is...

Jun's question via email about Runge Kutta Scheme...]]>

Take the Laplace Transform of the equation:

$\displaystyle \begin{align*} s\,Y\left( s \right) - y\left( 0 \right) + 11\,Y\left( s \right) &= \frac{3}{s^2} \\

s\,Y\left( s \right) - 5 + 11\,Y\left( s \right) &= \frac{3}{s^2} \\

\left( s + 11 \right) Y\left( s \right) &= \frac{3}{s^2} + 5 \\

Y\left( s \right) &= \frac{3}{s^2 \,\left(...

Adam's question via email about Laplace Transforms]]>

Upon taking the Laplace Transform of the equation we have

$\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &=...

Jun's question via email about Laplace Transform]]>

Taking the Laplace Transform of the equation gives

$\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right)...

Massaad's question via email about Laplace Transforms]]>

This requires the convolution theorem:

$\displaystyle \int_0^t{f\left( u \right) \,g\left( t- u \right) \,\mathrm{d}u } = F\left( s \right) \,G\left( s \right) $

In this case, $\displaystyle g\left( t - u...

Mahesh's question via email about Laplace Transforms (2)]]>

Start by taking the Laplace Transform of both equations, which gives

$\displaystyle...

Mahesh's question via email about Laplace Transforms (1)]]>

The Heaviside function suggests a second shift, but to do that, the entire function needs to be a function of $\displaystyle t - 4$.

Let $\displaystyle u = t - 4 \implies t = u + 4$, then

$\displaystyle \begin{align*} \mathrm{e}^{5\,t} &= \mathrm{e}^{5\left( u + 4 \right) } \\ &= \mathrm{e}^{5\,u + 20} \\

&= \mathrm{e}^{5\left( t - 4...

Alexander's question via email about Laplace Transforms]]>

$\displaystyle \begin{align*}

\mathcal{L} \left\{ 5\sin{ \left( 11\,t \right) } \sinh{ \left( 11\,t \right) } \right\} &= \mathcal{L} \left\{ 5\sin{ \left( 11\,t \right) } \cdot \frac{1}{2} \left( \mathrm{e}^{11\,t} - \mathrm{e}^{-11\,t} \right) \right\} \\

&= \frac{5}{2} \,\mathcal{L} \left\{ \mathrm{e}^{11\,t} \sin{ \left( 11\,t \right) } -...

Dharshan's question via email about a Laplace Transform]]>

In this case, I have not posted a link there.]]>

I have posted a link there to this topic so the OP can see my work.]]>

Since...

Jason's calculus questions]]>

$\displaystyle \begin{align*} F \left( \omega \right) &= \mathcal{F} \left\{ f \left( t \right) \right\} \\

&= \int_{-\infty}^{\infty}{ f\left( t \right) \mathrm{e}^{-\mathrm{j}\,\omega \, t}\,\mathrm{d}t } \\

&= \int_{-\infty}^{-2}{...

Erin's question via email about a Fourier Transform]]>

1. The graphs intersect where the functions are equal, so

$\displaystyle \begin{align*} 6 - x^2 &= 3 - 2\,x \\ 0 &= x^2 - 2\,x - 3 \\ 0 &= \left( x - 3...

Raj's integration questions via Facebook]]>

As all the z coefficients are the same, it's a good idea to eliminate the z values in the second and third equations, so apply R2 - R1 to R2 and R3 - R1 to R3...

$\displaystyle \begin{align*} z &= 12 - x + 4\,y \\ 0 &= -8 + 6\,x - y \\ 0 &= -7 - 11\,x - 9\,y \end{align*}$

Now we...

Mason's question via Facebook about solving a system of equations (2)]]>

The LCM of the $\displaystyle \begin{align*} x \end{align*}$ coefficients is 30, so multiplying the first equation by 6, the second by 10 and the third by 5 gives

$\displaystyle \begin{align*} 30\,x - 12\,y + 6\,z &= 18 \\ 30\,x + 10\,y + 30\,z &= 50 \\ 30\,x + 5\,y - 20\,z &=...

Mason's question via Facebook about solving a system of equations]]>

Harrison's question via Facebook about polar functions]]>

To start with, we need to realise that each modulus function will be defined differently depending on the value of $\displaystyle \begin{align*} x \end{align*}$.

Notice that

$\displaystyle \begin{align*} \left| x - 4 \right| = \begin{cases} x - 4 \textrm{ if } x \geq 4 \\ 4 - x \textrm{ if } x < 4 \end{cases}...

Alexander's problem via Facebook about solving a modulus equation]]>

We first need to remember that two lines are perpendicular when their gradients multiply to give -1...

Sameer's derivative problem via Facebook]]>

This equation is separable...

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}&= 3\,\sqrt{4 - y^2} \\ \frac{1}{\sqrt{4 - y^2}}\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 3 \\ \int{ \frac{1}{\sqrt{4 - y^2}}\,\frac{\mathrm{d}y}{\mathrm{d}x} \,\mathrm{d}x} &= \int{ 3\,\mathrm{d}x} \\ \int{...

Eugene's question via Facebook about a Differential Equation]]>

(a) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X < 3 \right) = \textrm{Pr}\,\left( Z < a \right) \end{align*}$, so if $\displaystyle \begin{align*} x = 3 \end{align*}$ and $\displaystyle \begin{align*} z = a \end{align*}$ then we have

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ a &= \frac{3 - 5}{2} \\ a &= \frac{-2}{\phantom{-}2} \\ a &= -1 \end{align*}$

(b) We are told $\displaystyle \begin{align*}...

James' question about Normal Distribution]]>

Since it's a PDF, that means the entire area under the curve must be 1, so

$\displaystyle \begin{align*} \int_0^1{ a \left( x^2 + b \right) \,\mathrm{d}x } &= 1 \\ a \left[ \frac{x^3}{3} + b\,x \right] _0^1 &= 1 \\ a \left[ \left( \frac{1^3}{3} + b\cdot 1 \right) - \left( \frac{0^3}{3} + b \cdot 0 \right) \right] &= 1 \\ a \left( \frac{1}{3} + b \right) &= 1 \\ a &= \frac{1}{ \frac{1}{3}+ b } \\ a &= \frac{3}{1 + 3\,b} \end{align*}$

Now as we are...

James' question about a continuous probability distribution]]>

1. By applying the sine rule we have

$\displaystyle \begin{align*} \frac{\sin{ \left( A \right) }}{a} &= \frac{\sin{ \left( B \right) } }{b} \\ \frac{\sin{ \left( \alpha \right) }}{10} &= \frac{\sin{ \left( \beta \right) }}{8} \\ \frac{\frac{2}{3}}{10} &= \frac{\sin{ \left( \beta \right) }}{8} \\ \frac{1}{15} &= \frac{\sin{ \left( \beta \right) }}{8} \\ \sin{ \left( \beta \right) } &= \frac{8}{15} \end{align*}$

2. The length of the diagonal must be...

Violet's Trigonometry Questions via Facebook]]>

(a) Since a = 2, that means $\displaystyle \begin{align*} \Delta x = \frac{2}{5} \end{align*}$

(b)

$\displaystyle \begin{align*} f \left( \frac{7\,\Delta x}{2} \right) &= f \left( \frac{7}{5} \right) \\ &= \left( \frac{7}{5} \right) ^2 \\ &= \frac{49}{25} \end{align*}$

(c)

$\displaystyle \begin{align*} A_4 &= f \left( \frac{7}{5} \right) \cdot \Delta x \\ &= \frac{49}{25} \cdot \frac{2}{5} \\ &= \frac{98}{125}...

Dale's questions via Facebook about Riemann Sums]]>

We should...

Edin's question via email about volume by revolution.]]>

Alagendram's question via email about approximating change]]>

This requires using Integration By Parts twice...

$\displaystyle \begin{align*} I &= \int{\mathrm{e}^{-2\,x}\cos{ \left( 3\,x \right) } \,\mathrm{d}x} \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} - \int{ -\frac{2}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)}\,\mathrm{d}x } \\ I &= \frac{1}{3}\,\mathrm{e}^{-2\,x} \sin{(3\,x)} + \frac{2}{3} \int{ \mathrm{e}^{-2\,x}...

Berk's question via email about an antiderivative]]>

Berk's question via email about approximating change]]>

(a) Differentiate both sides of the equation with respect to x:

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ y^3 + y + x\,y^2 \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ 10 + 4\sin{(x)} \right] \\ 3\,y^2\,\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{\mathrm{d}y}{\mathrm{d}x} +...

Edin's question via email about implicit differentiation]]>

I am assuming that this line integral is along the straight line from $\displaystyle \begin{align*} (0,0,0) \end{align*}$ to...

Luca's question via email about a line integral...]]>

Code:

```
//ReadData.cpp
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <sstream>
#include <iomanip>
#include <stdlib.h>
using namespace std;
int main() {
//Define the required variables
string title, head1, head2, s...
```

I have posted a link there to this question so the OP can view my work.]]>

It's not entirely obvious what to do with this question, as the denominator does not easily factorise. However, if we realise that $\displaystyle \begin{align*} s^4 + 40\,000 = \left( s^2 \right) ^2 + 200^2 \end{align*}$ it's possible to do a sneaky completion of the square...

$\displaystyle \begin{align*} \left( s^2 \right) ^2 + 200^2 &= \left( s^2 \right) ^2 + 400\,s^2 + 200^2 - 400\,s^2 \\ &= \left( s^2 + 200 \right) ^2 - 400\,s^2 \\ &= \left( s^2...

Henry's question via email about an Inverse Laplace Transform]]>

As the Heaviside function is a function of t - 4, that means all other terms must also be functions of t - 4. The sine function is, but the exponential isn't. However with a little manipulation, we get

$\displaystyle \begin{align*} f\left(...

Collin's question via email about a Laplace Transform]]>

Taking the Laplace Transform of both sides we have

$\displaystyle \begin{align*} \mathcal{L}\,\left\{ y'' + 4\,y \right\} &= \mathcal{L}\,\left\{ \mathrm{H}\,\left( t - 7 \right) \right\} \\ s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) +...

Collin's question via email about solving a DE using Laplace Transforms]]>

As the denominator is a function of s + 3, it suggests a shift had to have been utilised. As such, we also need the numerator to be a function of s + 3...

Let $\displaystyle \begin{align*} u = s + 3 \end{align*}$, then $\displaystyle \begin{align*} s = u-3 \end{align*}$ and thus

$\displaystyle \begin{align*} 5\,s^2 + 20\,s + 26 &=...

Collin's questions via email about Inverse Laplace Transforms]]>

$\displaystyle \begin{align*} A\,A^T &= \left[\begin{matrix} 3 & 0 & -4 \\ 4 & 0 & \phantom{-}3 \\ 0 & 5 & \phantom{-}0...

Sava's question via email about matrix multiplication]]>

A matrix is symmetric if it is equal to its own transpose, so to show $\displaystyle \begin{align*} C^T\,C \end{align*}$ is symmetric, we need to prove that $\displaystyle \begin{align*} \left( C^T\,C \right) ^T = C^T\,C \end{align*}$...

Sava's question via email about symmetric matrices]]>

As there is a repeated root, the partial fraction decomposition we should use is:

$\displaystyle \begin{align*} \frac{A}{x - 1} + \frac{B}{\left( x - 1 \right) ^2 } + \frac{C}{x - 2} &\equiv \frac{x^2}{\left( x - 1 \right) ^2\,\left( x - 2 \right) } \\ \frac{A\,\left( x - 1 \right) \left( x - 2 \right) + B\,\left( x - 2 \right) + C\,\left( x -...

Sava's question via email about integration with partial fractions.]]>

We should note that we can write any complex number as $\displaystyle \begin{align*} z = r\,\mathrm{e}^{\mathrm{i}\,\theta} \end{align*}$ where $\displaystyle \begin{align*} r = \left| z \right| \end{align*}$ and $\displaystyle \begin{align*} \theta = \textrm{arg}\,\left( z \right) + 2\,\pi\,n , \,\, n \in...

Question via email about complex numbers]]>

First let's write this number in its polar form.

$\displaystyle \begin{align*} \left| z \right| &= \sqrt{\left( -2 \right) ^2 + 2^2} \\ &= \sqrt{4 + 4} \\ &= \sqrt{8} \\ &= 2\,\sqrt{2} \end{align*}$

and as the number is in Quadrant 2

$\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) &= \pi - \arctan{ \left|...

Effie's question via email about Complex Numbers]]>

$\displaystyle \begin{align*} z^3 + 1 &= 0 \\ z^3 &= -1 \\ z^3 &= \mathrm{e}^{ \left( 2\,n + 1 \right) \,\pi\,\mathrm{i} } \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( 2\,n + 1 \right) \, \pi \,\mathrm{i}} \right] ^{\frac{1}{3}} \\ &= \mathrm{e}^{ \frac{\left( 2\,n + 1 \right) \,\pi}{3} \,\mathrm{i} } \end{align*}$

For the three solutions with $\displaystyle \begin{align*} \textrm{arg}\,\left( z...

Sava's question via email about solving complex number equations]]>

To start with, we should find the points of intersection of the two functions, as these will be the terminals of our regions of integration.

$\displaystyle \begin{align*} 2\,x^2 &= x + 1 \\ 2\,x^2 - x - 1...

Effie's question via email about a volume by revolution]]>

Q5. Here is a graph of the region to be integrated and the line to be rotated around.

First we should find the x intercept of the function $\displaystyle \begin{align*} y = 3 - 4\,\sqrt{x} \end{align*}$ as this will be the ending point of our region of integration.

$\displaystyle \begin{align*} 0 &= 3 - 4\,\sqrt{x} \\ 4\,\sqrt{x} &= 3 \\ \sqrt{x} &= \frac{3}{4} \\ x &= \frac{9}{16} \end{align*}$

To do this...

Brenton's questions via email about volume by revolution]]>

Here is a sketch of the region to be rotated around the y axis.

You first need to visualise this entire region being rotated around the y axis, to get a mental picture of what the solid looks like. Then you need to imagine that the solid is made up of very thin vertically-oriented hollow cylinders. You can then approximate the volume of the solid by adding up the volumes of all the cylinders.

The curved surface...

Divanshu's question via email about cylindrical shells]]>

Here is a graph of the region to be rotated. Notice that it is being rotated around the same line that is the lower boundary.

The volume will be exactly the same if everything is moved down by 4 units, with the advantage of being rotated around the x-axis. So using the rule for finding the volume of a solid formed by rotating $\displaystyle \begin{align*} f(x) \end{align*}$ around the x axis: $\displaystyle...

Divanshu's question via email about a volume by revolution]]>

Here is a sketch of the region to be rotated and the line to be rotated around.

Notice that the volume will be exactly the same if we were to move everything up by 3 units, but with the advantage of rotating around the x axis. So we want to find the volume of the region under $\displaystyle \begin{align*} y = 12 - x^2 \end{align*}$ between $\displaystyle \begin{align*} x= 0 \end{align*}$ and $\displaystyle...

Jesse's question via email about volume by revolution]]>

Here is a sketch of the region to be rotated.

To find a volume using cylindrical shells, you first need to picture what the region would like like when that area is rotated around the y axis. Then consider how it would look if that solid was made up of very thin cylinders.

Each cylinder has a curved area that is a rectangle (rotated around to form the cylinder), which has area $\displaystyle \begin{align*}...

Nour's question via email about volume of revolution using cylindrical shells.]]>

Here is a sketch of the region R and the line to be rotated around.

Clearly the x-intercept of $\displaystyle \begin{align*} y = 3 - 3\,\sqrt{x} \end{align*}$ is (1, 0) so the terminals of the integral will be $\displaystyle \begin{align*} 0 \leq x \leq 1 \end{align*}$.

We should note that the volume will be exactly the same if everything is moved up by 2 units, with the advantage of being rotated around the...

Kivindu's question via email about volume by revolution]]>

We should first find the $\displaystyle \begin{align*} x \end{align*}$...

Kishan's question via email about volume by revolution]]>

5. To start with, we should work out the x intercepts, they are x = -2 and x = 2. That means your region in the first quadrant will be integrated over $\displaystyle \begin{align*} x \in [0,2] \end{align*}$.

You should note that rotating the function $\displaystyle \begin{align*} y = 4 - x^2 \end{align*}$ about the line $\displaystyle \begin{align*} y = -\frac{1}{2} \end{align*}$ will give the exact same volume as rotating $\displaystyle...

Jake's questions about integrations]]>

We should use Partial Fractions to simplify the integrand. The denominator can be factorised further as $\displaystyle \begin{align*} \int{ \frac{54\,t - 12}{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } \end{align*}$, so that means the decomposition we should use is

$\displaystyle \begin{align*}...

Kishan's question via email about an indefinite integral]]>

8. Newton's Method for solving a nonlinear equation $\displaystyle \begin{align*} f(x) = 0 \end{align*}$ is $\displaystyle \begin{align*} x_{n+1} = x_n - \frac{f\left( x_n \right)}{f'\left( x_n \right) } \end{align*}$.

Here $\displaystyle \begin{align*} f(x) = 0.7\,x^4 + x - 4.75 = 0 \end{align*}$, so $\displaystyle \begin{align*} f'(x) = 2.8\,x^3 + 1 \end{align*}$. That means

$\displaystyle \begin{align*} x_3 &= x_2 - \frac{f(x_2)}{f'(x_2)} \\ &=...

Abhishek's Questions About Solving Nonlinear Equations]]>

One way to do this is to apply the product rule. To do this, we need to know the derivative of each factor.

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \, \left\{ \left[ \sinh{(7\,t)} \right] ^3 \right\} &= 7 \cdot \cosh{( 7\,t )} \cdot 3\left[ \sinh{(7\,t)} \right] ^2 \\ &= 21\cosh{(7\,t)}\left[ \sinh{(7\,t)} \right] ^2...

Ross' question via email about a derivative.]]>

$\displaystyle \begin{align*} y &= 15\log{ \left| \sec{ \left( 9\,t \right) } + \tan{ \left( 9\,t \right) } \right| } \\ &= 15 \log{ \left| \frac{1}{\cos{\left( 9\,t \right) } } + \frac{\sin{ \left( 9\,t \right) }}{\cos{ \left( 9\,t \right) }} \right| } \\ &= 15 \log{ \left| \frac{1 + \sin{\left( 9\,t \right) }}{\cos{...

3600244's question via email about a derivative]]>

$\displaystyle \begin{align*} \int{ 50\,t\cos{\left( 5\,t^2 \right) } \,\mathrm{d}t } &= 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) }\,\mathrm{d}t } \end{align*}$

Let $\displaystyle \begin{align*} u = 5\,t^2 \implies \mathrm{d}u = 10\,t\,\mathrm{d}t \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) }...

Effie's question via email about an indefinite integral.]]>

Since we have this relationship between x and y, as the two sides are equal, so are their derivatives. We just have to remember that as y is a function of x, any function of y is also a function of x, with the inner function "y" composed inside whatever is being told to do to the y. So to differentiate these parts the Chain Rule would be needed. All other rules like the product and quotient rules will still apply as well. Anyway, differentiating both...

Kamal's Questions via email about Implicit Differentiation]]>

To perform implicit differentiation we must make use of the chain rule. Basically if you have a function composed in another function, its derivative is the product of the inner function's derivative and the outer function's derivative. All other rules (such as the sum rule, the product rule, the quotient rule) still apply also.

We must also realise that any combination of "y" functions are essentially a function of x, as y is a function of x. Thus...

Effie's question via email about Implicit Differentiation]]>

We start by writing the system as an augmented matrix

$\displaystyle \begin{align*} \left[ \begin{matrix} 2 & -2 & 1 & 1 & \phantom{-}4 \\ 4 & -1 & 2 & 3 & -\frac{5}{2} \\ 4 & -4 & 5 & 1 & \phantom{-}8 \\ 0 & \phantom{-}2 & 3 & 4 & -5 \end{matrix} \right] \end{align*}$

Once we have used Gaussian Elimination to upper triangularise the system, the right hand column ends up becoming the elements of $\displaystyle \begin{align*} \mathbf{g} \end{align*}$...

James' question about Gaussian Elimination]]>

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 & 4 & \phantom{-}1 & 0 & \phantom{-}5 \\ \phantom{-}1 & 6 & -1 & 4 & \phantom{-}7 \\ -1 & 2 & -9 & 2 & -9 \\ \phantom{-}0 & 1 & \phantom{-}2 & 0 & \phantom{-}4 \end{matrix} \right] \end{align*}$

In order to solve the system we must upper triangularise the matrix. As we do, the right hand column becomes matrix $\displaystyle \begin{align*} \mathbf{g} \end{align*}$.

As the elements...

Sufyan's question via email about solving a system with Gaussian Elimination and Partial Pivoting]]>

Jamal's Q via email solving a system]]>

Jamal's question via email about solving a system with a PLU decomposition.]]>

The closest Inverse Laplace Transform from my table is

$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{2\,a\,s\,\omega}{\left( s^2 + \omega ^2 - a^2 \right) ^2 + 4\,a^2\,\omega ^2 } \right\} = \sin{ \left( \omega \, t \right) } \sinh{ \left( a \, t \right) } \end{align*}$

so we would...

Emad's question via email about Inverse Laplace Transform]]>

Effie's question via email about Eigenvalues, Eigenvectors and Diagonalisation]]>

This system can be written as a matrix equation $\displaystyle \begin{align*} A\,\mathbf{x} = \mathbf{b} \end{align*}$ as

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}2 & 4 & \phantom{-}0 & 1 \\ \phantom{-}2 & 8 & -2 & 7 \\ -2 & 2 & -2 & 7 \\ \phantom{-}0 & 8 & -5 & 11 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] = \left[ \begin{matrix} \phantom{-}7 \\ \phantom{-}3 \\ -7 \\ -12 \end{matrix}...

Question from Jesse about Gaussian Elimination and LU factorisation.]]>

I'm not sure where you're getting the idea that z = 2, as this is not correct.

I'm assuming this is to be done...

Question from Kamal about Gaussian Elimination via email.]]>

$$I = \int^{\pi}_{-\pi} \frac{x^4\cos(x)}{1-\sin(x)+\sqrt{1+\sin^2(x)}}dx$$]]>

I have posted a link there to this...

Phyllis' question at Yahoo! Answers regarding finding the line that divides an area into equal parts]]>

First, because the series is positive term, we don't have to worry about absolute values. Now $\displaystyle \begin{align*} a_n = \frac{2n + 3}{4n^3 + n} \end{align*}$ and

$\displaystyle \begin{align*} a_{n + 1} &= \frac{2\left( n + 1 \right) + 3}{4 \left( n + 1 \right) ^3 + n + 1} \\ &= \frac{2n + 2 + 3}{4 \left(...

Yonglie's question via email about ratio test]]>

To start with, apply partial fractions...

$\displaystyle \begin{align*} \frac{A\,s + B}{ s^2 + 9} + \frac{C\,s + D}{ \left( s^2 + 9 \right) ^2 } &\equiv \frac{ 4s^3 + 5s^2 + 57s + 45}{ \left( s^2 + 9 \right) ^2 } \\ \left( A\,s + B \right) \left( s^2 + 9 \right) + C\,s + D &\equiv 4s^3 + 5s^2 + 57s + 45 \\ A\,s^3 + 9A\,s + B\,s^2 +...

Johnsy's question over Facebook about an Inverse Laplace Transform]]>

Samantha's question at Yahoo! Answers involving related rates]]>

The first unsolved question is 'easy enough' and was posted on www.mathhelpforum.com by the user zokomoko...

Unsolved analysis and number theory from other sites...]]>

The big clue here is the square root in the denominator, because $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{x} \right) = \frac{1}{2\,\sqrt{x}} \end{align*}$. So this suggests that you probably need to find a square root function to substitute. Rewrite your integrand as

$\displaystyle \begin{align*} \int{ \frac{1}{\sqrt{5x}...

Johnsy's question about finding an indefinite integral]]>

To do this we should use implicit differentiation. If $\displaystyle \begin{align*} y = \arccot{(x)} \end{align*}$ then

$\displaystyle \begin{align*} \cot{(y)} &= x \\ \frac{\cos{(y)}}{\sin{(y)}} &= x \\ \frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{\cos{(y)}}{\sin{(y)}} \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left( x \right) \\...

Johnsy's question about finding a derivative via Facebook]]>

Posted on 11 26 2011 on www.artofproblemsolving.com by the member mr10123 and not yet solved…

Daniel plays a game with a random number generator. It randomly selects an... |

Unsolved statistics questions from other sites, part II]]>

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \sqrt{ \frac{1 - y^2}{1 - x^2} } \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\sqrt{ 1 - y^2 }}{\sqrt{1 - x^2} } \\ \frac{1}{\sqrt{1 - y^2}} \, \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\sqrt{1 - x^2} } \\ \int{ \frac{1}{\sqrt{1 - y^2}}\...

Johnsy's question via Facebook]]>

Here we will use the following transforms: $\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{n!}{ \left( a + \mathrm{i}\,\omega \right) ^{n+1} } \right\} = t^n\,\mathrm{e}^{-a\,t}\,\mathrm{H}(t) \end{align*}$ and $\displaystyle \begin{align*}...

Muhammed's question via email about an Inverse Fourier Transform (2)]]>

Muhammed's question via email about solving a DE using Fourier Transforms]]>

Completing the square gives

$\displaystyle \begin{align*} \frac{2\mathrm{i}\,\omega}{\omega ^2 + 10\omega + 29} &= \frac{2\mathrm{i}\,\omega}{ \omega ^2 + 10\omega + 5^2 - 5^2 + 29} \\ &= \frac{2\mathrm{i}\,\omega}{ \left( \omega + 5 \right) ^2 + 4 } \\ &= \frac{2\mathrm{i}\,\omega}{ \left( \omega + 5 \right) ^2 + 2^2 } \\ &= \frac{...

Muhammad's question via email about an Inverse Fourier Transform]]>

The first thing you need to do is take the Fourier Transform of both sides of the equation.

$\displaystyle \begin{align*} \mathcal{F} \left\{ f(t)...

Mohamed's question via email about solving an integral equation using Fourier transforms]]>

The line starting at $\displaystyle \begin{align*} \left( 0, 0, 0 \right) \end{align*}$ and ending at $\displaystyle \begin{align*} \left( \frac{1}{3}, \frac{\pi}{2}, 1 \right)...

Stefan's question about a line integral.]]>

I have posted a link there to this thread so the OP can view my work.]]>mangoqueen54 said:

I have posted a link there so the OP can view my work.]]>David said:

Here is the question:

I have posted a link there to this thread so the OP can view my work.]]>☺ said:

I have posted a link there to this thread so the OP can view my work.]]>Legend Of ~Incredim said:

The Royal Primate's question at Yahoo Answers regarding convergence of a series]]>The Royal Primate at Yahoo Answers said:

I have posted a link there to this thread so the OP can view my work.]]>Scott said:

I have posted a link there to this question so the OP can view my work.]]>