Author Topic: Mathematics Question Thread (Read 1626938 times) Tweet Share

RuiAce · « **Reply #2190 on:** July 17, 2017, 06:36:24 pm »

Quote from: sssona09 on July 17, 2017, 06:28:31 pm

This is rlly dumb but.. how would you exactly know cot (-135)

Do we see tan? And how do we use our knowledge of quadrants

Thank you

Assuming degrees, although not explicitly stated.
$\begin{align*}\cot (-135^\circ)&=\frac{1}{\tan (-135^\circ)}\\ &= -\frac{1}{\tan 135^\circ}\tag{tan odd function}\\ &=-\frac{1}{-\tan 45^\circ} \tag{2nd quad angle}\\ &=1\end{align*}$

georgiia · « **Reply #2191 on:** July 17, 2017, 07:16:14 pm »

Im having trouble with d

And this one

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RuiAce · « **Reply #2192 on:** July 17, 2017, 07:28:02 pm »

Quote from: georgiia on July 17, 2017, 07:16:14 pm

And this one

$PV = 500\implies P=\frac{500}{V}$
$\text{Differentiating ordinarily, }\frac{dP}{dv}=-\frac{500}{V^2}\\ \text{so when }V=10, \frac{dP}{dV}=-5$
Please provide progress on the previous parts as they are necessary for part d).

1937jk · « **Reply #2193 on:** July 18, 2017, 08:21:59 am »

Just stuck with this particle question, any help would be appreciated

A particle is accelerating according to the equation a = (3t + 1)² . if the particle is initially at rest 2 m to the left of 0, find its displacement after 4 secs.

RuiAce · « **Reply #2194 on:** July 18, 2017, 08:30:37 am »

Quote from: 1937jk on July 18, 2017, 08:21:59 am

Just stuck with this particle question, any help would be appreciated
A particle is accelerating according to the equation a = (3t + 1)² . if the particle is initially at rest 2 m to the left of 0, find its displacement after 4 secs.

$v=\frac{(3t+1)^3}{9}+C_1\\ \text{When }t=0,v=0\\ \text{so }v=\frac{(3t+1)^3}{9}-\frac19$
$x=\frac{(3t+1)^4}{108}-\frac{t}{9}+C_2\\ \text{When }t=0, x=-2\\ \text{so }C_2 =-\frac{217}{108}\\ \text{and }x=\frac{(3t+1)^4}{108}-\frac{t}{9}-\frac{217}{108}$
$\text{Simply plug in }t=4.$

itssona · « **Reply #2195 on:** July 18, 2017, 08:58:33 am »

Pls help

Rewrite the following as a positive acute angle:
Cos 195

I know its -cos(theta) but how can we get a positive answer? :/

RuiAce · « **Reply #2196 on:** July 18, 2017, 09:04:02 am »

Quote from: sssona09 on July 18, 2017, 08:58:33 am

Pls help
Rewrite the following as a positive acute angle:
Cos 195

I know its -cos(theta) but how can we get a positive answer? :/

$\text{If that's degrees, }\cos 195^\circ\text{ is certainly negative.}$

itssona · « **Reply #2197 on:** July 18, 2017, 09:11:53 am »

Quote from: RuiAce on July 18, 2017, 09:04:02 am

$\text{If that's degrees, }\cos 195^\circ\text{ is certainly negative.}$

Ah thank you! Glad it wasn't a problem with my lack of understanding xD

jaysun · « **Reply #2198 on:** July 18, 2017, 03:21:46 pm »

Hey, could I please get some help with question b) both parts... I'm really confused how to go about this; my teacher taught this really badly and Im super scared for trials!!!

Thank youu!!!!

sudodds · « **Reply #2199 on:** July 18, 2017, 03:28:26 pm »

Quote from: jaysun on July 18, 2017, 03:21:46 pm

Hey, could I please get some help with question b) both parts... I'm really confused how to go about this; my teacher taught this really badly and Im super scared for trials!!!

Thank youu!!!!

Hi Jaysun,

so for (i) I was thinking you would differentiate (xe^x) and get xe^x + e^x which would be your first mark.

For (ii):
d/dx(xe^x) = e^x+xe^x
d/dx(xe^x) - e^x = xe^x
then 'integral' d/dx (xe^x) - e^x.dx = 'integral' xe^x.dx
finally: xe^x - e^x + C = 'integral' xe^x.dx

For this just try to make the second part resemble the first bit - so you see how part (i) had an extra e^x, you just manipulate the equation in part (ii) to resemble the earlier one.

Hope this helps!

Susie

Shadowxo · « **Reply #2200 on:** July 18, 2017, 03:35:21 pm »

a) So the product rule is $\frac{d}{dx}(u*v)=u*\frac{dv}{dx}+v*\frac{du}{dx} \\ or \frac{d}{dx}(u*v)= u*v' +v*u' \text{ where ' indicates the derivative}$
So
$\frac{d}{dx}(x*e^x) = x* \frac{d}{dx}(e^x) + e^x * \frac{d}{dx}(x) \\ = x * e^x + e^x$
b) So we know $\frac{d}{dx}(x*e^x) = x * e^x + e^x \\ x*e^x = \frac{d}{dx}(x*e^x)-e^x \text{ so}\\ \int x*e^x = \int (\frac{d}{dx}(x*e^x)-e^x) = x*e^x - e^x + c$
Basically, you rearrange it, and also the integral of the derivative of something is just the thing itself. You rearrange the equation so that what you're integrating is the subject, and usually you end up with regular integrals or integral of d/dx(something). Hard to explain but examples help

Also sorry reply took a while, wanted to make it easy to read

RuiAce · « **Reply #2201 on:** July 18, 2017, 04:41:56 pm »

Quote from: sudodds on July 18, 2017, 03:28:26 pm

Hi Jaysun,

so for (i) I was thinking you would differentiate (xe^x) and get xe^x + e^x which would be your first mark.

For (ii):
d/dx(xe^x) = e^x+xe^x
d/dx(xe^x) - e^x = xe^x
then 'integral' d/dx (xe^x) - e^x.dx = 'integral' xe^x.dx
finally: xe^x - e^x + C = 'integral' xe^x.dx

For this just try to make the second part resemble the first bit - so you see how part (i) had an extra e^x, you just manipulate the equation in part (ii) to resemble the earlier one.

Hope this helps!

Susie

Thanks Isaac/Jamon

sudodds · « **Reply #2202 on:** July 18, 2017, 04:46:12 pm »

Quote from: RuiAce on July 18, 2017, 04:41:56 pm

Thanks Isaac/Jamon

You spelt Susie wrong

itssona · « **Reply #2203 on:** July 18, 2017, 05:33:27 pm »

Heey can u pls check if these are right:
Parabola:
Domain - all real x
Range- all real y

Log x
Domain - all positivr x
Range- all real y

Absolute value:
Range: all real y

Cubic:
Domain: all real x??
Range: all real y

Exponential :
Range: all positive y
Domain: all real x

RuiAce · « **Reply #2204 on:** July 18, 2017, 05:44:13 pm »

Quote from: sssona09 on July 18, 2017, 05:33:27 pm

Heey can u pls check if these are right:
Parabola:
Domain - all real x
Range- all real y

Log x
Domain - all positivr x
Range- all real y

Absolute value:
Range: all real y

Cubic:
Domain: all real x??
Range: all real y

Exponential :
Range: all positive y
Domain: all real x

Not sure what's going on with the red ones.
$\textbf{Also, standard curves }\text{e.g. }y=\ln x\textbf{ have been assumed}\\ \textbf{This is, in general, not okay}\\ \text{as something like }y=\ln (x+1)\text{ will have a different domain}$

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