Author Topic: VCE Methods Question Thread! (Read 5415144 times) Tweet Share

IndefatigableLover · « **Reply #4410 on:** April 07, 2014, 06:36:24 pm »

Quote from: jessss0407 on April 07, 2014, 06:04:01 pm

Hey guys! I need help with this question:
1a) For the functions f and g that are defined and differentiable for all real numbers, it is
known that:
f(1)=6, g(1)=−1, g(6)=7 and f(−1)=8
f'(1)=6, g'(1)=−2, f'(−1)=2 and g'(6)=−1
Find:
i) (f◦g)'(1)
ii) (g◦f)'(1)
iii) (fg)'(1)
iv) (gf)'(1)
v) (f/g)'(1)
vi) (g/f)'(1)

Thanks!

This question is really just using the chain/product/quotient rule but it's worded pretty weirdly ._.

i). Taking it a step back, if we know that 'f◦g' = f(gx(x)) then the derivative of 'f◦g' is equal to f'(g(x))g'(x) (to which we know as the chain rule) then we can simply substitute that into the equation!

$\therefore f'(g(1))g'(1)=f'(-1) \times -2 = -4$ (because g(1)=-1 and f'(-1) = 2)

ii). Pretty much the same as i) but swapping around the values.

g◦f'(1) $= g'(f(1))f'(1)= g'(6) \times 6 = -6$ (because f(1) = 6 and g'(6)=-1)

iii). This is where the product rule comes in! So you know that with the product rule it's pretty much:

(fg)'(1) = f'(1) x g(1) + g'(1) x f(1)

$\therefore f'(1)g(1) + g'(1)f(1) = 6 \times -1 + -2 \times 6 = -18$

iv). Same as before but just swapping it around (I'm sure you can input the values yourself for this one

)

v). (f/g)'(1) = Quotient Rule!

$(\frac{f}{g})'(x) = \frac{f'(x) \times g(x) - g'(x) \times f(x)}{[g(x)^2]}$

$\therefore (\frac{f}{g})'(1) = \frac{f'(1) \times g(1) - g'(1) \times f(1)}{[g(1)^2]}$

$= \frac{6 \times -1 -(-2 \times 6)}{[-1]^2}=6$

vi). Same as v) but swapping the values around.

Hope that helps (and if I got anything wrong then please do tell)!

(Sorry for the awks Latex but hopefully it's readable)

Anchy · « **Reply #4411 on:** April 07, 2014, 09:34:40 pm »

Having some trouble with this question:

Find the exact solution for x in e^2+x = (2e^1/2x)/3

Thanks

Phy124 · « **Reply #4412 on:** April 07, 2014, 09:48:10 pm »

Is that $e^{2+x} = \frac{2e^{\frac{1}{2}x}}{3}$ ?

Anchy · « **Reply #4413 on:** April 07, 2014, 09:51:43 pm »

Quote from: Phy124 on April 07, 2014, 09:48:10 pm

Is that $e^{2+x} = \frac{2e^{\frac{1}{2}x}}{3}$ ?

Yep

b^3 · « **Reply #4414 on:** April 07, 2014, 10:14:22 pm »

Quote from: Anchy on April 07, 2014, 09:34:40 pm

Having some trouble with this question:

Find the exact solution for x in e^2+x = (2e^1/2x)/3

Thanks

$\begin{alignedat}{1}e^{2+x} & =\frac{2}{3}e^{\frac{1}{2}x} \\ e^{2+\frac{1}{2}x} & =\frac{2}{3} \\ \text{"log\_base of the answer gives the power"} \\ 2+\frac{1}{2}x & =\log_{e}\left(\frac{2}{3}\right) \\ \frac{1}{2}x & =\log_{e}\left(\frac{2}{3}\right)-2 \\ \therefore x & =2\log_{e}\left(\frac{2}{3}\right)-4 \end{alignedat}$

wildareal · « **Reply #4415 on:** April 08, 2014, 02:34:15 pm »

Probability Question:

1. Given that a player has a Blackjack, what is the conditional probability the Banker will get a Blackjack?
2. What is the probability the player had a blackjack given the Banker got one? (Consider the total probability of getting Blackjack for the Banker and for the Player).
3. What is the probability both Player and Banker will get Blackjack?

wildareal · « **Reply #4416 on:** April 08, 2014, 02:42:30 pm »

^N.B: This is assuming that Player is dealt the first two cards from the pack and then the Banker is dealt the next two cards.

wildareal · « **Reply #4417 on:** April 08, 2014, 02:44:57 pm »

1. Pr(Player Blackjack)=(4/52*16/51)*2=32/663
Pr(Banker Blackjack Given Banker)=32/663*Banker Blackjack

Banker Blackjack=(3/50*14/49)*2=84/2450

therefore:

32/663*84/2450
=64/38675

jessss0407 · « **Reply #4418 on:** April 09, 2014, 09:14:07 am »

Quote from: IndefatigableLover on April 07, 2014, 06:36:24 pm

This question is really just using the chain/product/quotient rule but it's worded pretty weirdly ._.

i). Taking it a step back, if we know that 'f◦g' = f(gx(x)) then the derivative of 'f◦g' is equal to f'(g(x))g'(x) (to which we know as the chain rule) then we can simply substitute that into the equation!

$\therefore f'(g(1))g'(1)=f'(-1) \times -2 = -4$ (because g(1)=-1 and f'(-1) = 2)

ii). Pretty much the same as i) but swapping around the values.

g◦f'(1) $= g'(f(1))f'(1)= g'(6) \times 6 = -6$ (because f(1) = 6 and g'(6)=-1)

iii). This is where the product rule comes in! So you know that with the product rule it's pretty much:

(fg)'(1) = f'(1) x g(1) + g'(1) x f(1)

$\therefore f'(1)g(1) + g'(1)f(1) = 6 \times -1 + -2 \times 6 = -18$

iv). Same as before but just swapping it around (I'm sure you can input the values yourself for this one )

v). (f/g)'(1) = Quotient Rule!

$(\frac{f}{g})'(x) = \frac{f'(x) \times g(x) - g'(x) \times f(x)}{[g(x)^2]}$

$\therefore (\frac{f}{g})'(1) = \frac{f'(1) \times g(1) - g'(1) \times f(1)}{[g(1)^2]}$

$= \frac{6 \times -1 -(-2 \times 6)}{[-1]^2}=6$

vi). Same as v) but swapping the values around.

Hope that helps (and if I got anything wrong then please do tell)!

(Sorry for the awks Latex but hopefully it's readable)

Ohhhh thanks so much!

alchemy · « **Reply #4419 on:** April 09, 2014, 09:41:04 am »

What is the area of the set of points (x,y) defined by |2x−3y|≤12 and |2x+3y|≤12 ?

So, I think I've plotted this wrong in the first place, but here is my attempt: https://www.desmos.com/calculator/rl1r47crme
Calculating the area from that gives 24.

However, when I plot the graph exactly as the equation states, I get: https://www.desmos.com/calculator/vabllfawiv . The area is 48, which seems to be the correct answer even according to this: https://www.wolframalpha.com/input/?i=|2x-3y|%3C%3D12+and+|2x%2B3y|%3C%3D12#.

What's the mistake I'm making in plotting the first graph? I've used the method how I would normally sketch absolute value functions.

Zealous · « **Reply #4420 on:** April 09, 2014, 03:24:07 pm »

Quote from: alchemy on April 09, 2014, 09:41:04 am

What is the area of the set of points (x,y) defined by |2x−3y|≤12 and |2x+3y|≤12 ?

So, I think I've plotted this wrong in the first place, but here is my attempt: https://www.desmos.com/calculator/rl1r47crme
Calculating the area from that gives 24.

However, when I plot the graph exactly as the equation states, I get: https://www.desmos.com/calculator/vabllfawiv . The area is 48, which seems to be the correct answer even according to this: https://www.wolframalpha.com/input/?i=|2x-3y|%3C%3D12+and+|2x%2B3y|%3C%3D12#.

What's the mistake I'm making in plotting the first graph? I've used the method how I would normally sketch absolute value functions.

I've got a wonky method of solving this. So there are probably better ways of doing this.

$|2x-3y|\leq12 \hspace{3mm} and \hspace{3mm} |2x+3y|\leq12$

Let's find the intersection points of these two graphs, so we can find the area in between them. As the both equal 12, we can equate the graphs. We can express the modulus sign as the square root of something squared (in order to remove the negatives).
$|2x-3y|=|2x+3y| \rightarrow \sqrt{(2x-3y)^2}=\sqrt{(2x+3y)^2}$

Square both sides then squareroot both sides:
$(2x-3y)^2}=(2x+3y)^2} \rightarrow \pm(2x-3y)=\pm(2x+3y)$

Testing the positive of the plus or minus sign:
$2x-3y=3x+3y \rightarrow y=0$
That is these two graphs will intersect when y=0.

Testing the negative of the plus or minus sign:
$3y-2x=-2x-3y \rightarrow x=0$
The two graphs will also intersect when x=0.

Now we know that the two graphs will intersect when x=0 and when y=0. We can now sub these values into the original equations to find the coordinates of the intersection point.

Sub in x=0 into one of the original equations:
$\sqrt{(2(0)-3y)^2}=12 \rightarrow (-3y)^2=144 \rightarrow -3y=\pm12 \rightarrow y=\pm4$
So two of our intersection points are (0,4) and (0,-4)

Sub in y=0 into one of the original equations:
$\sqrt{(2x-3(0))^2}=12 \rightarrow (2x)^2=144 \rightarrow 2x=\pm12 \rightarrow y=\pm6$
So the other two intersection points are (6,0) and (-6,0)

Our four intersection points are (6,0), (-6,0), (0,4) and (0,-4) which can be clearly seen in what you linked previously: https://www.desmos.com/calculator/vabllfawiv. You can then plot these points on a cartesian plane, connect them with lines (as the absolute function in this case is linear) and find the area in between using simple geometry with triangles; the area is 48 square units.

jessss0407 · « **Reply #4421 on:** April 09, 2014, 06:07:54 pm »

Hi guys!

I have a few questions
1. Find the derivative of x²(x²+2x)^1/3

I got x²(x²+2x)^-2/3(8x/3 + 8/3)
whereas the answers had $2x(x^{2}+2x)^{1/3}+\frac{2x^{2}(x+1)}{3(x^{2}+2x)^{2/3}}$
so i'm not really sure how to check if my answer is also correct

2. Find the derivative of (x²+1)(x³+1)^1/3

3. Can someone explain why the derivative of log_enx is always 1/x regardless of what n is?

Thanks

Zealous · « **Reply #4422 on:** April 09, 2014, 06:19:51 pm »

Quote from: jessss0407 on April 09, 2014, 06:07:54 pm

3. Can someone explain why the derivative of log_enx is always 1/x regardless of what n is?

Thanks

3: Using the chain rule, you can see that n will cancel out with n on the denominator.
$y=log_{e}{(nx)} , \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$

let u=nx

$y=log_e{(u}) \rightarrow \frac{dy}{du}=\frac{1}{u}=\frac{1}{nx}$

$u=nx \rightarrow \frac{du}{dx}=n$

$\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}=\frac{1}{nx} \times n =\frac{1}{x}$

Sorry ran out of time to type up responses to all 3 of your questions - got to eat haha! hopefully that helps q3.

IndefatigableLover · « **Reply #4423 on:** April 09, 2014, 06:35:50 pm »

Quote from: jessss0407 on April 09, 2014, 06:07:54 pm

Hi guys!

I have a few questions
1. Find the derivative of x²(x²+2x)^1/3

I got x²(x²+2x)^-2/3(8x/3 + 8/3)
whereas the answers had $2x(x^{2}+2x)^{1/3}+\frac{2x^{2}(x+1)}{3(x^{2}+2x)^{2/3}}$
so i'm not really sure how to check if my answer is also correct

2. Find the derivative of (x²+1)(x³+1)^1/3

3. Can someone explain why the derivative of log_enx is always 1/x regardless of what n is?

Thanks

Question 1: (Latex)

School taught me to do things this way but I'll just step you through it!

$x^2.(x^2+2x)^\frac{1}{3}$

So you know you're using the product rule (basically u'v + v'u)
Because we have: u, u', v, v', let's get those out of the way first!

$Let \ u = x^2$

$Let \ u' = 2x$

$Let \ v = (x^2+2x)^\frac{1}{3}$

$Let \ v' = \frac{2}{3}(x+1)(x^2+2x)^\frac{-2}{3}$

And so then we know that:

$\frac{d}{dx} = u'v+v'u$

Then sub it in!

$= (2x)(x^2+2x)^\frac{1}{3}+ \frac{2}{3}(x+1)(x^2+2x)^\frac{-2}{3}(x^2)$

$= (2x)(x^2+2x)^\frac{1}{3}+ \frac{2}{3}x^2(x+1)(x^2+2x)^\frac{-2}{3}$

And then we just rearrange it to make it look like the answer you've given!

$\therefore = (2x)(x^2+2x)^\frac{1}{3}+ \frac{2x^2(x+1)}{3(x^2+2x)^\frac{2}{3}}$

That's the answer they've stated but you could go more in depth if you wanted to as well (like expanding the last one and factoring a bit)! I think a major problem if differentiation is knowing when to stop simplifying I guess...

Normally when I check my answer for these questions I just use Wolfram Alpha to check (iPad app)

jessss0407 · « **Reply #4424 on:** April 09, 2014, 07:14:45 pm »

Quote from: Zealous on April 09, 2014, 06:19:51 pm

3: Using the chain rule, you can see that n will cancel out with n on the denominator.
$y=log_{e}{(nx)} , \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$

let u=nx

$y=log_e{(u}) \rightarrow \frac{dy}{du}=\frac{1}{u}=\frac{1}{nx}$

$u=nx \rightarrow \frac{du}{dx}=n$

$\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}=\frac{1}{nx} \times n =\frac{1}{x}$

Sorry ran out of time to type up responses to all 3 of your questions - got to eat haha! hopefully that helps q3.

ahah no worries! thanks so much

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