How to do this mc question? (1 Viewer)

malcolm21 · Oct 24, 2015

what is amplitude and period of: v^2 = 16(9-x^2)

Crisium · Oct 24, 2015

It's in the form v^2 = n^2 (a^2 - x^2)

So

Amplitude = 3

Period = 2 pi / 4

= pi / 2

malcolm21 · Oct 24, 2015

Crisium said:
It's in the form v^2 = n^2 (a^2 - x^2)

So

Amplitude = 3

Period = 2 pi / 4

= pi / 2

thanks man im a dummy

malcolm21 · Oct 24, 2015

Can someone help me with this question too:

Use the binomial theorem to find an expression for the constant term
in the expansion of (2x^3 - 1/x)^12

Im very weak with the last two topics motion and binomial

turnerloos · Oct 24, 2015

Tk+1 = (n k) (2x^3)^12-k (-x^-1)^k
Find coeff of x^0

InteGrand · Oct 24, 2015

malcolm21 said:
Can someone help me with this question too:

Use the binomial theorem to find an expression for the constant term
in the expansion of (2x^3 - 1/x)^12

Im very weak with the last two topics motion and binomial

$Note that$

$\left( 2x^3 -\frac{1}{x}\right)^{12} = \left( 2x^3 -x^{-1}\right)^{12}$

$= \sum _{k=0}^{12} \binom{12}{k}\left(2x^3\right)^{12-k}\left(-x^{-1} \right)^k$

$= \sum _{k=0}^{12} \binom{12}{k}2^{12-k}x^{36-3k}\cdot (-1)^k \cdot x^{-k}$

$$= \sum _{k=0}^{12} \binom{12}{k}(-1)^k 2^{12-k}x^{36-4k}$ (index laws).$

$The general term is thus $T_k = \binom{12}{k}(-1)^k 2^{12-k}x^{36-4k}$, for $k=0,1,2,\ldots, 12$. constant term has a power of $0$ for $x^{36-4k}$, so constant term has $36-4k=0\Rightarrow k = 9$. So the constant term is $T_9 = \binom{12}{9}(-1)^9 2^{12-9}= -\binom{12}{9}\times 2^3 = -1760$.$

kawaiipotato · Oct 24, 2015

turnerloos said:
Tk+1 = (n k) (2x^3)^12-k (-x^-1)^k
Find coeff of x^0

Or alternatively collect the fraction together
= (2x^4 - 1)^12 / x^12
and use the general term to find x^12 at the top as x^12 / x^12 gives the constant term

SNT k · Oct 24, 2015

They probably wont ask that though, most likely a binominal proof of some sort.

Crisium · Oct 24, 2015

SNT k said:
They probably wont ask that though, most likely a binominal proof of some sort.

They often ask it in the MC section

turnerloos · Oct 24, 2015

kawaiipotato said:
Or alternatively collect the fraction together
= (2x^4 - 1)^12 / x^12
and use the general term to find x^12 at the top as x^12 / x^12 gives the constant term

alternative pathways are always a +1

malcolm21 · Oct 24, 2015

one more question, how does the equation change like this:

turnerloos · Oct 24, 2015

auxiliary angle

kawaiipotato · Oct 24, 2015

malcolm21 said:
one more question, how does the equation change like this:

You can rewrite the addition/subtraction of either sine/cosine as one trigonometric function.
$6 \cos 2t + 8 \sin 2t \equiv R \cos (2t - \alpha)$

$\equiv R \cos2t \cos \alpha + R \sin 2t \sin \alpha$

$Equating,$

$6 = R \cos \alpha$

$8 = R \sin \alpha$

$Squaring and adding,$

$8^2 + 6^2 = 100 = R^2$

$R = 10$

$\therefore 6 \cos 2t + 8 \sin 2t = 10 \cos (2t - \alpha)$

$($You can turn it into a sine function too but they chose cosine and simply found the 10 by just squaring and adding then square rooting$)$

InteGrand · Oct 24, 2015

malcolm21 said:
one more question, how does the equation change like this:

$This is the auxiliary angle transformation. We can write $6\cos 2t + 8\sin 2t \equiv R\cos (2t-\alpha)$, for some $R>0$ and $\alpha \in \mathbb{R}$. Expanding with the compound angle formula, we have$

$$6\cos 2t + 8\sin 2t \equiv R \cos 2t \cos \alpha + R\sin 2t \sin \alpha$,$

$and equating coefficients of $\cos 2t$ and $\sin 2t$ gives us the following system of equations:$

$\left\{\begin{matrix}R\cos \alpha = 6 \ \ (1)\\ R\sin \alpha =8 \ \ (2)\end{matrix}\right.$

$Squaring and adding, we have $R^2 (\cos^2 \alpha + \sin^2 \alpha) = 6^2 + 8^2 \Rightarrow R^2 \times 1 = 10^2 \Rightarrow R = 10$, as $R>0$.$

$Now, Equations (1) and (2) tell us that $\alpha$ is in quadrant 1 and Equation (1) tells us that $10\cos \alpha = 6\Rightarrow \cos \alpha =\frac{3}{5}\Rightarrow \alpha = \cos^{-1}\left(\frac{3}{5} \right)$ does the trick.$

$So $6\cos 2t + 8\sin 2t \equiv 10\cos \left(2t-\cos^{-1}\left(\frac{3}{5} \right)\right) $.$

$(Edit: done above. This has how to find $\alpha$ if you're interested.)$

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