Author Topic: VCE Specialist 3/4 Question Thread! (Read 2171438 times) Tweet Share

Fishyiscool · « **Reply #375 on:** April 01, 2012, 02:37:04 pm »

Two questions:

If sin A = 1/sqrt5, how do i find sin 3A?

and

how to find exact values of sin (pi/8) using double angle formulas?

pi · « **Reply #376 on:** April 01, 2012, 02:57:56 pm »

Quote from: tisaraiscool on April 01, 2012, 02:37:04 pm

Two questions:

If sin A = 1/sqrt5, how do i find sin 3A?

and

how to find exact values of sin (pi/8) using double angle formulas?

Hints only!

1) Find sin(2A) using relevant formula and then find sin(3A) using the sin(X+Y) formulas where X=A and Y=3A

2) Let 2A = pi/4, use sin(2A) identity to solve for sin(A)

Bhootnike · « **Reply #377 on:** April 03, 2012, 01:17:35 pm »

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}$ $sec^2x tan^5x\, dx$

evaluate:

can somsone please guide me through that!

Ominous · « **Reply #378 on:** April 03, 2012, 01:56:43 pm »

Quote from: Bhootnike on April 03, 2012, 01:17:35 pm

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}$ $sec^2x tan^5x\, dx$

evaluate:

can somsone please guide me through that!

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}$ $sec^2x tan^5x\, dx$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{2}x \cdot \tan^{5}x \cdot dx=$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+tan^{2}(x)) \cdot \tan^{5}x \cdot dx$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+tan^{2}(x)) \cdot \tan^{5}x \cdot dx$

$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+u^{2}) \cdot u^{5} \cdot dx$

let u= $1+ \tan^{2} x$ , also change up the terminals and use u-subsitution $\int_{\frac{3+\sqrt{3}}{3}}^{1+\sqrt{3}} u^{5} \cdot u^{7} du$

and every would be straight forward.

EDIT: its just a matter of using the trig identites of $\sec^{2}x= 1+tan^2{x}$ . and changing terminals through u subsitution.

Good luck

Bhootnike · « **Reply #379 on:** April 03, 2012, 06:10:06 pm »

woah,
so ... i got as far as you did in the 3rd /4th line.

you've then let tan = u by the looks of it, but then said

Quote from: Ominous on April 03, 2012, 01:56:43 pm

let u= $1+ \tan^{2} x$

b^3 · « **Reply #380 on:** April 03, 2012, 06:20:17 pm »

Ill just help you with the the intergal, then you can sub in the values yourself later.
So we have $\int\sec^{2}(x)\tan^{5}(x)dx$
Now here there is something we can notice, we know that the derivative of $\tan(x)$ is $\sec^{2}(x)$ , so if we let u=tan(x), then we can try and get the $\sec^{2}(x)$ to cancel out.

$\mathrm{Let\: u=\tan(x)}$
$\mathrm{Then\:}\frac{du}{dx}=\sec^{2}(x)$
So we have $\begin{alignedat}{1}\int\sec^{2}(x)\tan^{5}(x)dx & =\intop u^{5}\sec^{2}(x)\times\frac{1}{\sec^{2}(x)}du \\ & =\int u^{5}du \\ & =\frac{1}{6}u^{6}+c \\ & =\frac{1}{6}\tan^{6}(x)+c \end{alignedat}$

Then sub in the values to find the definite integral.
Oh, and don't forget to change the terminals

Bhootnike · « **Reply #381 on:** April 03, 2012, 06:24:58 pm »

Quote from: b^3 on April 03, 2012, 06:20:17 pm

Ill just help you with the the intergal, then you can sub in the values yourself later.
So we have $\int\sec^{2}(x)\tan^{5}(x)dx$
Now here there is something we can notice, we know that the derivatie of $\tan(x)$ is $\sec^{2}(x)$ , so if we let u=tan(x), then we can try and get the $\sec^{2}(x)$ to cancel out.

$\mathrm{Let\: u=\tan(x)}$
$\mathrm{Then\:}\frac{du}{dx}=\sec^{2}(x)$
So we have $\begin{alignedat}{1}\int\sec^{2}(x)\tan^{5}(x)dx & =\intop u^{5}\sec^{2}(x)\times\frac{1}{\sec^{2}(x)}du \\ & =\int u^{5}du \\ & =\frac{1}{6}u^{6}+c \\ & =\frac{1}{6}\tan^{6}(U)+c \end{alignedat}$

Then sub in the values to find the definite integral.
Oh, and don't forget to change the terminals

beautttty. thanks!

nina_rox · « **Reply #382 on:** April 04, 2012, 10:16:13 am »

Question: what does the second derivative tell you about the original graph? I read up on it but I'm very confused about concavity stuff? Thanks!

pi · « **Reply #383 on:** April 04, 2012, 01:56:28 pm »

Quote from: nina_rox on April 04, 2012, 10:16:13 am

Question: what does the second derivative tell you about the original graph? I read up on it but I'm very confused about concavity stuff? Thanks!

It 'usually' tells you (there are ambiguous exceptions, such as y = x^4) whether or not a particular point (based on the value of x) is a local max/min or a certain type of inflection

Bhootnike · « **Reply #384 on:** April 04, 2012, 03:40:15 pm »

Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$ = $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$ ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

pi · « **Reply #385 on:** April 04, 2012, 04:04:19 pm »

Quote from: Bhootnike on April 04, 2012, 03:40:15 pm

Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$ = $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$ ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

In general $\int^a_b f(x) \ dx = -\int^b_a f(x) \ dx = \int^b_a -f(x) \ dx$

In your case, the negative has been moved inside the integral

Planck's constant · « **Reply #386 on:** April 04, 2012, 04:06:14 pm »

Quote from: VegemitePi on April 04, 2012, 04:04:19 pm

Quote from: Bhootnike on April 04, 2012, 03:40:15 pm
Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$ = $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$ ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

In general $\int^a_b f(x) \ dx = -\int^b_a f(x) \ dx = \int^b_a -f(x) \ dx$

In your case, the negative has been moved inside the integral

10/10

Bhootnike · « **Reply #387 on:** April 04, 2012, 04:23:10 pm »

Quote from: VegemitePi on April 04, 2012, 04:04:19 pm

Quote from: Bhootnike on April 04, 2012, 03:40:15 pm
Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$ = $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$ ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

In general $\int^a_b f(x) \ dx = -\int^b_a f(x) \ dx = \int^b_a -f(x) \ dx$

In your case, the negative has been moved inside the integral

ahhh indeed! ok

thanks!
handy lil tip right there!

nina_rox · « **Reply #388 on:** April 04, 2012, 06:45:10 pm »

Quote from: VegemitePi on April 04, 2012, 01:56:28 pm

Quote from: nina_rox on April 04, 2012, 10:16:13 am
Question: what does the second derivative tell you about the original graph? I read up on it but I'm very confused about concavity stuff? Thanks!

It 'usually' tells you (there are ambiguous exceptions, such as y = x^4) whether or not a particular point (based on the value of x) is a local max/min or a certain type of inflection

Could you explain this further? We got a worksheet where there were a few exercise to do on this and one of the question was 'if you draw a tangent on f(x) curve at the point f''(x)=0 what does it do and tell us?' Thank you for your help!

ashoni · « **Reply #389 on:** April 04, 2012, 07:04:46 pm »

Resolve the following rational expressions into partial fractions.
(3x^2+2x+5) / (x^2+2)(x+1)
AND
(x^2+2x-13) / (2x^3+6x^2+2x+6)

show workingg please.. thanks!

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2171438 times) Tweet Share

Fishyiscool

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