Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

RuiAce · « **Reply #1095 on:** April 18, 2017, 10:16:01 pm »

Quote from: Rathin on April 18, 2017, 10:06:50 pm

Help with this please
∫sin(x)sin(3x)dx

$\text{To derive a useful product-to-sum formula}\\ \begin{align*}\cos(A-B)-\cos(A+B)&= 2\sin A \sin B\end{align*}$
$\text{So taking }A=3x, \, B=x\\ \int \sin 3x\sin x\, dx = \frac12\int (\cos 2x - \cos 4x)dx$

Rathin · « **Reply #1096 on:** April 19, 2017, 09:53:08 am »

Quote from: RuiAce on April 18, 2017, 10:16:01 pm

$\text{To derive a useful product-to-sum formula}\\ \begin{align*}\cos(A-B)-\cos(A+B)&= 2\sin A \sin B\end{align*}$
$\text{So taking }A=3x, \, B=x\\ \int \sin 3x\sin x\, dx = \frac12\int (\sin 2x - \sin 4x)dx$

Hey, wouldn't sin(x)sin(3x)=1/2(cos(2x)-cos(4x))?

RuiAce · « **Reply #1097 on:** April 19, 2017, 09:56:55 am »

Quote from: Rathin on April 19, 2017, 09:53:08 am

Hey, wouldn't sin(x)sin(3x)=1/2(cos(2x)-cos(4x))?

No idea why I wrote down sin again. Let me edit that

Rathin · « **Reply #1098 on:** April 20, 2017, 01:10:55 pm »

Quote from: RuiAce on April 19, 2017, 09:56:55 am

No idea why I wrote down sin again. Let me edit that

Textbook answer says something else from my answer. I got 1/4(sin(2x)) - 1/8(sin(4x)) +c . Is this correct?

Rathin · « **Reply #1099 on:** April 20, 2017, 01:27:53 pm »

By IBP I got I=(sin(3x)cos(x)-3cos(3x)sin(x))/8 +c which is what the textbook has as its answer.

jakesilove · « **Reply #1100 on:** April 20, 2017, 02:59:52 pm »

Quote from: Rathin on April 20, 2017, 01:27:53 pm

By IBP I got I=(sin(3x)cos(x)-3cos(3x)sin(x))/8 +c which is what the textbook has as its answer.

Those two solutions are exactly the same

RuiAce · « **Reply #1101 on:** April 20, 2017, 04:46:40 pm »

What Jake said. When integrating, you must ALWAYS consider the possibility that the final answers are no different to each other.

crazycodpro · « **Reply #1102 on:** April 20, 2017, 07:35:20 pm »

RuiAce · « **Reply #1103 on:** April 20, 2017, 07:47:15 pm »

Quote from: crazycodpro on April 20, 2017, 07:35:20 pm

$\text{This is nothing different to 3U related rates.}$
$\text{Let the basal diameter and height be }h\\ \text{Note then that the basal radius is }\frac{h}{2}$
$\text{The volume of the cone formed is given by }V=\frac{1}{3}{\pi r^2 h}\\ \text{Since }r=\frac{h}{2}\text{ we have }V=\frac{\pi h^3}{12}\\ \implies \frac{dV}{dh}= \frac{\pi h^2}{4}$
$\text{Using the chain rule}\\ \frac{dV}{dt}=\frac{dV}{dh} \, \frac{dh}{dt}\\ \text{Substituting everything in}\\20 = \frac{\pi \times 8^2}{4} \, \frac{dh}{dt}$

crazycodpro · « **Reply #1104 on:** April 20, 2017, 08:15:44 pm »

this is a stupid question but RuiAce how did you find the Dv/dh

RuiAce · « **Reply #1105 on:** April 20, 2017, 08:16:17 pm »

Quote from: crazycodpro on April 20, 2017, 08:15:44 pm

this is a stupid question but RuiAce how did you find the Dv/dh

Just differentiate it.
$\frac{d}{dx}x^n = nx^{n-1}$
$\frac{d}{dx}x^3=3x^2$

crazycodpro · « **Reply #1106 on:** April 20, 2017, 08:20:16 pm »

that make sense, thanks RuiAca

beau77bro · « **Reply #1107 on:** April 21, 2017, 01:46:07 pm »

Please help hahaha

RuiAce · « **Reply #1108 on:** April 21, 2017, 02:19:32 pm »

beau77bro · « **Reply #1109 on:** April 23, 2017, 12:13:29 am »

Please help me question 4c, I feel I've done it before, but a solution would be so helpful

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