Author Topic: Mathematics Question Thread (Read 1626908 times) Tweet Share

kylesara · « **Reply #2070 on:** June 17, 2017, 02:26:06 pm »

Hi could I please have help with this question
The sum of the first 3 terms of an A.P is 33 and the sum of the seventh and eighth term is 49. Find the first term and common difference.
Thanks!

RuiAce · « **Reply #2071 on:** June 17, 2017, 02:29:44 pm »

Quote from: kylesara on June 17, 2017, 02:26:06 pm

Hi could I please have help with this question
The sum of the first 3 terms of an A.P is 33 and the sum of the seventh and eighth term is 49. Find the first term and common difference.
Thanks!

$S_3 = 33 \implies \boxed{\frac{3}{2}\left[2a+2d\right]=33}\\ \text{For convenience this can be simplified to }\boxed{a+d=11}$
$\text{Then we just use the formula for the nth term: }T_n = a+(n-1)d\\ T_7+T_8 = \boxed{a+6d+a+7d = 49}$
$\text{This is now a system of simultaneous equations}\\ \text{The solution is }\boxed{a=\frac{94}{11}, d=\frac{27}{11}}$

katnisschung · « **Reply #2072 on:** June 19, 2017, 03:08:57 pm »

stuck on b thanks

jakesilove · « **Reply #2073 on:** June 19, 2017, 03:14:23 pm »

Quote from: katnisschung on June 19, 2017, 03:08:57 pm

stuck on b thanks

$\frac{dy}{dx}=\tan^2x\ sec^2x -sec^2 x+1$

We turn everything into tans
$=\tan^2x(\tan^2x+1)-(\tan^2x+1)+1$
$=\tan^4x+tan^2x-tan^2x-1+1$

Which gets to our answer

katnisschung · « **Reply #2074 on:** June 19, 2017, 03:14:52 pm »

need help in differentiating this

my answer... i need help in simplifying this further i think
q & ans attached
my answer... also i swear i remember my teacher telling me never to expand/simplify
the denominator for quotient rule is this true?

jakesilove · « **Reply #2075 on:** June 19, 2017, 03:25:41 pm »

Quote from: katnisschung on June 19, 2017, 03:14:52 pm

need help in differentiating this

my answer... i need help in simplifying this further i think
q & ans attached
my answer... also i swear i remember my teacher telling me never to expand/simplify
the denominator for quotient rule is this true?

$\frac{d}{dx}(-(\cos x+\sin x)^{-2})$
$=2(\cos x + \sin x)^{-3}*(-\sin x + \cos x)$

Using the chain rule. No real need to simplify further.

$\frac{d}{dx}\frac{\cos x}{\cos x + \sin x}$
$=\frac{(\cos x + \sin x)(-\sin x)-(\cos x)(-\sin x+\cos x)}{(\cos x + \sin x)^2}$
using Quotient rule
$=\frac{-1}{(\cos x +\sin x)^2}$

By trig identities. Remember that
$\sin^2 x+ \cos^2 x=1$

katnisschung · « **Reply #2076 on:** June 19, 2017, 03:30:10 pm »

thanks jake

seems that trig identities keep tripping me up. time to memorise them

jakesilove · « **Reply #2077 on:** June 19, 2017, 03:32:54 pm »

Quote from: katnisschung on June 19, 2017, 03:30:10 pm

thanks jake seems that trig identities keep tripping me up. time to memorise them

Definitely need to memorise them! At least, memorise the above one, because you can derive all the others from there

katnisschung · « **Reply #2078 on:** June 20, 2017, 07:04:44 pm »

so can somebody pls explain to me how this is not integrating by the y-axis...
i looked at the answers and it just integrates by the x-axis but isn't the area
bounded by the y-axis?

yeah i see that my perspective is flawed cos i can't see any way to rearrange the first equation in terms of x
but the area is bounded by the y-axis!!

confused. hope someone can make sense out of this.

RuiAce · « **Reply #2079 on:** June 20, 2017, 07:45:29 pm »

Quote from: katnisschung on June 20, 2017, 07:04:44 pm

so can somebody pls explain to me how this is not integrating by the y-axis...
i looked at the answers and it just integrates by the x-axis but isn't the area
bounded by the y-axis?

yeah i see that my perspective is flawed cos i can't see any way to rearrange the first equation in terms of x
but the area is bounded by the y-axis!!
confused. hope someone can make sense out of this.

$\text{If you wanted to integrate it by considering the }y\text{-axis}\\ \text{You would have a compound region and be forced to deal with this}\\ A=\int_0^1 y\, dy + \int_1^{\sqrt{2}}\frac{4}{\pi}\cos^{-1}\frac{y}{\sqrt2} \, dy\\ \text{And when you see the second integral you realise you ran into problems}$
$\textbf{Just because the }y\textbf{-axis bounds it does NOT mean you should analyse it w.r.t. }y\\ \textbf{You need to always consider the possibility of area between curves}\\ \text{By considering area }\textbf{between}\text{ curves, you no longer care if any axes bounds it.}$
$\text{The easier to tackle integral is }A=\int_0^2 \left(\sqrt{2}\cos \frac{\pi x}{4} - x\right)dx$

itssona · « **Reply #2080 on:** June 21, 2017, 09:20:08 am »

Heey
how would i graph y=5cos4x

I know amplitude is 5 and the period is pi/2 but i dont get how to label the x axis in terms of radians and then graph it

Thank you!

RuiAce · « **Reply #2081 on:** June 21, 2017, 09:27:45 am »

Quote from: sssona09 on June 21, 2017, 09:20:08 am

Heey
how would i graph y=5cos4x

I know amplitude is 5 and the period is pi/2 but i dont get how to label the x axis in terms of radians and then graph it

Thank you!

Sketch the typical cosine curve, but then label your max and min values as 5 and -5 instead of 1 and -1. Then, you can just label the x-axis as normal

cos(x) has period 2pi, intercepts π/2,3π/2,...
So noting the period of Pi/2 our intercepts are π/8,3π/8,...

Aussie1Italia2 · « **Reply #2082 on:** June 21, 2017, 09:40:18 am »

Hello,
I need help sketching for 0<= x <= PI

y=7sin^x/₂

Thank you for the help.

MisterNeo · « **Reply #2083 on:** June 21, 2017, 09:41:53 am »

Quote from: sssona09 on June 21, 2017, 09:20:08 am

Heey
how would i graph y=5cos4x

I know amplitude is 5 and the period is pi/2 but i dont get how to label the x axis in terms of radians and then graph it

Thank you!

Consine graphs like 5cos4x start at the amplitude value on the positive y-axis, which would be 5.

The first time it touches the x-axis will be (pi/8,0). After that, you repeat with the Pi/2 periods.
The easy way to graph is to remember:

Get the coefficient of the x (4) and multiply it by 2, then divide pi by it.
That is your first x-value for a cosine graph.

Hope this helps

RuiAce · « **Reply #2084 on:** June 21, 2017, 09:51:50 am »

Quote from: Aussie1Italia2 on June 21, 2017, 09:40:18 am

Hello,
I need help sketching for 0<= x <= PI

y=7sin^x/₂

Thank you for the help.

This question is very similar to the one posted right before it, so all I'm gonna do is copy and paste the response and change numbers.
_______

Sketch the typical sine curve, but then label your max and min values as 7 and -7 instead of 1 and -1. Then, you can just label the x-axis as normal

sin(x) has period 2π, intercepts 0,π,2π,...
So noting the period of 4π our intercepts are 0,2π,4π,...

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