Author Topic: VCE Methods Question Thread! (Read 6267502 times) Tweet Share

soNasty · « **Reply #4815 on:** May 19, 2014, 08:34:09 pm »

can someone tell me how to anti differentiate:

$\frac{3x}{10log_e(\frac{1}{4})}$

Zealous · « **Reply #4816 on:** May 19, 2014, 08:47:12 pm »

Quote from: soNasty on May 19, 2014, 08:34:09 pm

can someone tell me how to anti differentiate:

$\frac{3x}{10log_e(\frac{1}{4})}$

$\begin{eqnarray*}\\\int(\frac{3x}{10log_{e}(\frac{1}{4})})dx & = & \frac{3}{10log_{e}(\frac{1}{4})}\int(x)dx\\& = & \frac{3}{10log_{e}(\frac{1}{4})}(\frac{x^{2}}{2})+c=\frac{3x^{2}}{20log_{e}(\frac{1}{4})}+c\\\end{eqnarray*}\\\\$

If you wanna go even further (using logarithm laws):

$\frac{3x^{2}}{20log_{e}(\frac{1}{4})}+c & = & \frac{3x^{2}}{20log_{e}(2^{-2})}+c=-\frac{3x^{2}}{40log_{e}(2)}+c\\$

soNasty · « **Reply #4817 on:** May 19, 2014, 08:56:06 pm »

oh my god yes thats right you can take that whole thing out!!
ahaha thanks zealous

Jason12 · « **Reply #4818 on:** May 20, 2014, 12:17:26 pm »

how to derive

sin^2(x-pi/4)

sin^2(x)

Zealous · « **Reply #4819 on:** May 20, 2014, 12:27:31 pm »

Quote from: Jason12 on May 20, 2014, 12:17:26 pm

how to derive

sin^2(x-pi/4)

sin^2(x)

$\begin{eqnarray*}\\let\ y & = & sin^{2}(x)=sin(x)^{2}\\\frac{dy}{dx} & = & 2sin(x)cos(x)\\\end{eqnarray*}$

Confused? Chain rule:

$\begin{eqnarray*}\\\frac{dy}{dx} & = & \frac{dy}{du}\times\frac{du}{dx}\\\\let\ u & = & sin(x)\rightarrow\frac{du}{dx}=cos(x)\\y=sin(x)^2 & = & u^{2}\rightarrow\frac{dy}{du}=2u\\\\\therefore\frac{dy}{dx} & = & 2u\times cos(x)=2sin(x)cos(x)\\\end{eqnarray*}$

Both of your questions are just applications of the chain rule. Hope this helps!

b^3 · « **Reply #4820 on:** May 20, 2014, 12:29:15 pm »

I think you mean differentiate?

Hint: Use the chain rule.

Spoiler

$\begin{alignedat}{1}y & =\sin^{2}\left(x-\frac{\pi}{4}\right) \\ \text{Let }u= & \sin\left(x-\frac{\pi}{4}\right) \\ y & =u^{2} \\ \frac{dy}{dx} & =\frac{dy}{du}\cdot\frac{du}{dx} \\ \frac{dy}{du} & =2u \\ \frac{du}{dx} & =\cos\left(x-\frac{\pi}{4}\right) \\ \frac{dy}{dx} & =2u\cdot\cos\left(x-\frac{\pi}{4}\right) \\ & =2\sin\left(x-\frac{\pi}{4}\right)\cos\left(x-\frac{\pi}{4}\right) \\ & =\sin\left(2\left(x-\frac{\pi}{4}\right)\right) \\ & =\sin\left(2x-\frac{\pi}{2}\right) \end{alignedat}$

The second can be done similarily.

Edit: Beaten.

Jason12 · « **Reply #4821 on:** May 20, 2014, 01:01:05 pm »

Quote from: b^3 on May 20, 2014, 12:29:15 pm

I think you mean differentiate?

Hint: Use the chain rule.
Spoiler
$\begin{alignedat}{1}y & =\sin^{2}\left(x-\frac{\pi}{4}\right) \\ \text{Let }u= & \sin\left(x-\frac{\pi}{4}\right) \\ y & =u^{2} \\ \frac{dy}{dx} & =\frac{dy}{du}\cdot\frac{du}{dx} \\ \frac{dy}{du} & =2u \\ \frac{du}{dx} & =\cos\left(x-\frac{\pi}{4}\right) \\ \frac{dy}{dx} & =2u\cdot\cos\left(x-\frac{\pi}{4}\right) \\ & =2\sin\left(x-\frac{\pi}{4}\right)\cos\left(x-\frac{\pi}{4}\right) \\ & =\sin\left(2\left(x-\frac{\pi}{4}\right)\right) \\ & =\sin\left(2x-\frac{\pi}{2}\right) \end{alignedat}$

The second can be done similarily.

Edit: Beaten.

thanks but i don't get the last bit where you do 2sin(x-pi/4) x cos(x-pi/4)

Zealous · « **Reply #4822 on:** May 20, 2014, 01:11:06 pm »

Quote from: Jason12 on May 20, 2014, 01:01:05 pm

thanks but i don't get the last bit where you do 2sin(x-pi/4) x cos(x-pi/4)

It's a double angle formula, you don't need to know it for Methods (you use it a fair bit in Specialist though).

Jason12 · « **Reply #4823 on:** May 20, 2014, 01:46:22 pm »

just to confirm is this correct?

$cos^2(x)= cos(x)^2 = [cos(x)]^2$

Reus · « **Reply #4824 on:** May 20, 2014, 03:43:33 pm »

Quote from: Jason12 on May 20, 2014, 01:46:22 pm

just to confirm is this correct?

(Image removed from quote.)

I know that the first two are...

Stevensmay · « **Reply #4825 on:** May 20, 2014, 04:19:56 pm »

Quote from: Jason12 on May 20, 2014, 01:46:22 pm

just to confirm is this correct?

(Image removed from quote.)

Yes all are equivalent.

Jason12 · « **Reply #4826 on:** May 20, 2014, 05:32:57 pm »

If I have h(x) defined as something on a ti-nspire cx cas how can I undefine it without deleting the whole document?

byCrypt · « **Reply #4827 on:** May 20, 2014, 05:57:41 pm »

Quote from: Jason12 on May 20, 2014, 05:32:57 pm

If I have h(x) defined as something on a ti-nspire cx cas how can I undefine it without deleting the whole document?

I don't think you can undefine it, but you can just define h(x) for something else and the previous one will be deleted.

Mieow · « **Reply #4828 on:** May 20, 2014, 06:07:29 pm »

I need help with two questions:

1)Find the equation of the quadratic curve, given that f(0) = 0 and f(4) = 0 and f ′(0) = -1.

2) Find the derivative of y= $log_{e}\tfrac{x+1}{x-1}$

Thanks!

nhmn0301 · « **Reply #4829 on:** May 20, 2014, 06:15:44 pm »

Quote from: Mieow on May 20, 2014, 06:07:29 pm

I need help with two questions:

1)Find the equation of the quadratic curve, given that f(0) = 0 and f(4) = 0 and f ′(0) = -1.

2) Find the derivative of y= $log_{e}\tfrac{x+1}{x-1}$

Thanks!

1) let y=ax^2 + bx + c be the function you want to find
f(0) = 0 means c=0
f(4) = 0 means 14a + 4b = 0
f ' (x) = 2ax + b (treats a,b as a constant)
f ' (0) = -1 means 2a(0) + b = -1 -> b=-1
You can do the rest from here, just sub in the value of b to find a.
b) you can re-write this equation as
y= ln (x+1) - ln (x-2)
dy/dx= 1/ (x+1) - 1/(x-2)

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