Author Topic: VCE Methods Question Thread! (Read 4874776 times) Tweet Share

Cort · « **Reply #5040 on:** June 08, 2014, 01:40:24 pm »

I've already done this question, but I just wanted to get some other insight into the thinking process involved with it.

Because Z(a,b) is basically touching, you can always assume that b = y value, yes? Because it's 'touching' ? Or is there another reason why to it?

Thanks,
Cort.

Orb · « **Reply #5041 on:** June 08, 2014, 01:49:47 pm »

Quote from: Cort on June 08, 2014, 01:40:24 pm

I've already done this question, but I just wanted to get some other insight into the thinking process involved with it.

Because Z(a,b) is basically touching, you can always assume that b = y value, yes? Because it's 'touching' ? Or is there another reason why to it?

Thanks,
Cort.

Yeah i'm pretty sure until you figure out the equation you assume this to be the case.

Billion · « **Reply #5042 on:** June 08, 2014, 02:16:58 pm »

Quote from: hamo94 on June 08, 2014, 01:30:46 pm

Inflation rate is the percentage that the value rises in accordance to time.
So inflation rate is dV/dt
= 0.02 + (5pi/6) x cos ((pi/6)t)

That's how I thought that you'd approach the question, but I'm not quite sure how you obtain the 0.1% per month value

It has something to do with 0.02/$20 x 100 = 0.1%, but I'm not sure why we're doing this.

Phy124 · « **Reply #5043 on:** June 08, 2014, 02:55:05 pm »

Quote from: Cort on June 08, 2014, 01:40:24 pm

I've already done this question, but I just wanted to get some other insight into the thinking process involved with it.

Because Z(a,b) is basically touching, you can always assume that b = y value, yes? Because it's 'touching' ? Or is there another reason why to it?

Thanks,
Cort.

"A rectangle XYZW has ... two vertices, Y and Z, on the graph of $y = 9-3x^2$ ..." means that both Y and Z are points on the graph $y=9-3x^2$ and that is why you can equate the two $y$ values.

Mieow · « **Reply #5044 on:** June 08, 2014, 03:20:25 pm »

Thank you so much Zealous!

#ZealousForPresident

Jason12 · « **Reply #5045 on:** June 08, 2014, 11:38:24 pm »

help with anti differentiating these

b^3 · « **Reply #5046 on:** June 08, 2014, 11:47:29 pm »

Expand the first two out before you try to integrate, then follow $\int x^{n} dx=\frac{1}{n+1}x^{n+1}+C,\: C\in\mathbb{R}$ .
$\begin{alignedat}{1}\frac{x}{x+1} & =\frac{x+1-1}{x+1} \\ & =\frac{x+1}{x+1}-\frac{1}{x+1} \\ & =1-\frac{1}{x+1} \\ \frac{x+1}{x} & =\frac{x}{x}+\frac{1}{x} \\ & =1+\frac{1}{x} \end{alignedat}$

For the third, as the coefficient in front of the $x$ is $1$ you can look at it just like if you were integrating $\frac{1}{x^{2}}$ , except with $x+1$ in it's place.

Spoiler

$\begin{alignedat}{1}\int\frac{x}{x+1}dx & =\int\left(1-\frac{1}{x+1}\right)dx \\ & =x-\log_{e}\left(|x+1|\right)+C,\: C\in\mathbb{R} \\ \int\frac{x+1}{x}dx & =\int\left(1+\frac{1}{x}\right)dx \\ & =x+\log_{e}\left(|x|\right)+C,\: C\in\mathbb{R} \\ \int\frac{1}{\left(x+1\right)^{2}}dx & =-\frac{1}{x+1}+C,\: C\in\mathbb{R} \end{alignedat}$

Jason12 · « **Reply #5047 on:** June 09, 2014, 12:40:08 am »

logs cant be negative right? and if they are do they become 0 in an equation?

ie. 10 = -loge^1 + c

c = 10

kinslayer · « **Reply #5048 on:** June 09, 2014, 12:54:07 am »

Quote from: Jason12 on June 09, 2014, 12:40:08 am

logs cant be negative right? and if they are do they become 0 in an equation?

ie. 10 = -loge^1 + c

c = 10

The logarithm (to any base) of 1 is zero. $\log_a 1 = 0$ for any a.

It's not clear what you've written down there. is it $10 = -\log_e 1 + c$ ? Then c = 10 because log(1) = 0.

The logarithm (to any base**) of a number is negative if that number is less than 1.*

Example: $\log_e (1/e) = \log_e (e^{-1}) = -\log_e e = -1$

* this is because $a^x < 1$ if and only if $x < 0$

** for a > 1

edit: I like b^3's explanation better than mine. Here is a graph link to explain/redeem myself:

https://www.desmos.com/calculator/obhs9ulszj (use the slider for a)

b^3 · « **Reply #5049 on:** June 09, 2014, 12:54:18 am »

The inside of a log can't be negative but you can have a negative log.
Lets look at the following.
$\begin{alignedat}{1}\text{Let }x & =\log_{e}\left(a\right) \\ \implies e^{x} & =a \end{alignedat}$
Now we know that $e^{x}$ will be positive no matter what power we raise it to, (if $x$ is large then $e^{x}$ is large, if $x$ is really small $e^{x}$ is really small).

Now I'm going to assume in your equation you have $\log_{e}(1)$ ?
Let's look at why any log of 1 we have is zero.
$\begin{alignedat}{1}\text{Let }x & =\log_{b}\left(1\right) \\ \implies b^{x} & =1 \\ \implies x & =0 \\ \implies\log_{b}\left(1\right) & =0 \end{alignedat}$
As the only power that will give one in this situation is zero.

EDIT: Beaten, 6 seconds at the initial attempt to post

Jason12 · « **Reply #5050 on:** June 09, 2014, 01:40:35 pm »

help with anti differentiating e^2x + 1/e^x

i think you can split the into e^2x/e^x and 1/e^x but can you cancel e^x and e^2x? this might be a silly question but just need confirmation

Zealous · « **Reply #5051 on:** June 09, 2014, 01:48:18 pm »

Quote from: Jason12 on June 09, 2014, 01:40:35 pm

help with anti differentiating e^2x + 1/e^x

i think you can split the into e^2x/e^x and 1/e^x but can you cancel e^x and e^2x? this might be a silly question but just need confirmation

$\begin{eqnarray*}\\\int(e^{2x}+\frac{1}{e^{x}}) \ dx & = & \int(e^{2x}+e^{-x}) \ dx\\ & = & \frac{1}{2}e^{2x}-e^{-x}+C\\\end{eqnarray*}\\$

Spoiler

$\int e^{ax}dx=\frac{1}{a}e^{ax}+C$
Pretty much the reverse of when you differentiate e (assuming a is a constant).

[edit] Was there supposed to be a bracket around the numerator? Then this:

$\\\begin{eqnarray*}\\\int(\frac{e^{2x}+1}{e^{x}})dx & = & \int(\frac{e^{2x}}{e^{x}}+\frac{1}{e^{x}})dx=\int(e^{x}+e^{-x})dx\\ & = & e^{x}-e^{-x}+C\\\end{eqnarray*}\\$

Jason12 · « **Reply #5052 on:** June 09, 2014, 02:37:22 pm »

$e^2=\frac{1}{2}e^2 + c$

how to solve for c? again should be simple but i keep getting it wrong. Answer says C = 1/2e^2

IndefatigableLover · « **Reply #5053 on:** June 09, 2014, 02:44:14 pm »

Quote from: Jason12 on June 09, 2014, 02:37:22 pm

(Image removed from quote.)

how to solve for c? again should be simple but i keep getting it wrong. Answer says C = 1/2e^2

Well $e^2$ is the same as $\frac{2e^2}{2}$ whilst $\frac{1}{2}e^2$ can be rewritten as $\frac{e^2}{2}$ so if you subtract that from both sides you should end up with the right answer

Jason12 · « **Reply #5054 on:** June 09, 2014, 03:16:25 pm »

how to do (-1/4 x (7)^-2) - (-1/4 x (3)^-2)

i don't think you can make (7)^-2 = 2/7 right?

also how to cancel loge^2/loge^6

and what is -1/2loge^3 + 1/2loge^9

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