Author Topic: VCE Methods Question Thread! (Read 5555957 times) Tweet Share

ashoni · « **Reply #331 on:** February 23, 2012, 07:29:23 pm »

Quote from: paulsterio on February 23, 2012, 07:21:25 pm

(Image removed from quote.)

hmm that is very strange.. well if it matters or not, the answer at the back says m= -6 +or- 4 sqrt(2)

Deceitful Wings · « **Reply #332 on:** February 25, 2012, 07:16:31 pm »

This was a question on a methods test done in school, could you please help?
The line 3y-x=6 is normal to the curve y=ax³ - 2
Find the exact value of a
[4 marks]

Phy124 · « **Reply #333 on:** February 25, 2012, 07:57:53 pm »

Hmmm I'm going to go with the following...

$3y - x =6$

$3y = x + 6$

$y = \frac{1}{3}x+2$

$\therefore m = \frac{1}{3}$

______

$y=ax^{3}-2$

$\frac{dy}{dx} = 3ax^{2}$

Because the equation is at a normal to the other the gradient at the intersection will be $-\frac{1}{m} = -3$ , so:

$3ax^{2} = -3$

______

We also know it will be at the point the lines intersect, so:

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

So we have two simultaneous equations

$3ax^{2} = -3$

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

Solve for x and a (shits about to get hectic):

$ax^{3}-2=\frac{x+6}{3}$

$3ax^{2}=-3$

$3ax^{3}-6=x+6$

$3ax^{2}=-3$

$3ax^{3}-x=12 (\times ax^{2})$

$3ax^2 = -3 (\times ax^{3})$

$3a^{2}x^{5}-ax^{3}=12ax^{2}$

$3a^{2}x^{5}=-3ax^{3}$

$-ax^{3}=12ax^{2}+3ax^{3}$

$-4ax^{3} = 12ax^{2}$

$-4x = 12$

$x = -3$

Sub x back into one of original equations:

$3a(-3)^{2}=-3$

$3a(9)=-3$

$27a = -3$

$a = -\frac{1}{9}$

Therefore, $a = -\frac{1}{9}$ and the point at which the normal will occur is $x = -3$

*Note: You don't have to solve for x, I just did that because they often ask when it will occur, in which case you would have to.

Could be way off, but hey, hope it's right

edit: Before I forget, it only looks really complicated because I didn't cancel out anything at the start and included every single step, always look to make your work as minimal as possible, or as much so without making mistakes (I just enjoy it more this way, yeah I said that...

)

edit2: Go with what trinh answered, probably easier to follow.

trinh · « **Reply #334 on:** February 25, 2012, 08:09:30 pm »

For the curve,
$y=ax^3-2$
$\frac{dy}{dx}=3ax^2$

For the normal,
$y=\frac{1}{3}x+2$

As Phy124 showed,
at $(x,y)$ :
$3ax^2=-3$
=> $ax^2=-1$ [equation 1]
and
$ax^3-2=\frac{1}{3}x+2$ [equation 2]

Now sub [1] into [2]:
$-1x-2=\frac{1}{3}x+2$
$x=-3$

Then sub $x=-3$ back into [1]:
$a(-3)^2=-1$
$a=\frac{-1}{9}$

Deceitful Wings · « **Reply #335 on:** February 25, 2012, 09:55:22 pm »

Thanks a bunch guys, just wandering if you could help with one more

Let g(x) be a function such that g(-2)=1/9 and g'(-2)=-5
If f(x)= sqrt(g(x)), find the simplest form the value of f'(-2).

i got something like 3/2 but i have a feeling it is wrong :S

trinh · « **Reply #336 on:** February 25, 2012, 10:20:04 pm »

$g(x):$
$g(-2)=\frac{1}{9}$
$g'(-2)=-5$

$f(x)=\sqrt(g(x))$
$(f(x))^2=g(x)$

So...
$g(-2)=\frac{1}{9}$ => $f(-2)=\frac{1}{3}$
and
$g'(-2)=-5$ => $2f(-2)f'(-2)=-5$ (using chain rule)

Using $2f(-2)f'(-2)=-5$ , we now know

$2*\frac{1}{3}f'(-2)=-5$
$f'(-2)=\frac{-15}{2}$

Now that I finish, I realise that you wanted f'(2); did you make a typo and actually need f'(-2), or did I just waste time? $:-\$

kamil9876 · « **Reply #337 on:** February 25, 2012, 10:25:11 pm »

Chain rule pretty much:

If $f(x)=h(g(x))$ then $f'(x)=g'(x) h'(g(x))$ so in our case, $h(y)=y^{\frac{1}{2}}$ so we get $f'(x)=g'(x) \frac{1}{2} g(x)^{-\frac{1}{2}$ and hence at x=-2 that is: $f'(-2) = -5 * \frac{1}{2} *\left( \frac{1}{9} \left) ^{-0.5}=\frac{-15}{2}$

kamil9876 · « **Reply #338 on:** February 25, 2012, 10:28:15 pm »

Quote

Now that I finish, I realise that you wanted f'(2); did you make a typo and actually need f'(-2), or did I just waste time? $:-\$

Oh right, good point. I think we can deduce that it is indeed a typo, because knowing the derivative of a function and it's value at a point does not uniquely determine how it behaves globally. In other words it would be impossible to give one answer, f'(2) could be anything.

Deceitful Wings · « **Reply #339 on:** February 25, 2012, 10:29:55 pm »

Quote from: trinh on February 25, 2012, 10:20:04 pm

$g(x):$
$g(-2)=\frac{1}{9}$
$g'(-2)=-5$

$f(x)=\sqrt(g(x))$
$(f(x))^2=g(x)$

So...
$g(-2)=\frac{1}{9}$ => $f(-2)=\frac{1}{3}$
and
$g'(-2)=-5$ => $2f(-2)f'(-2)=-5$ (using chain rule)

Using $2f(-2)f'(-2)=-5$ , we now know

$2*\frac{1}{3}f'(-2)=-5$
$f'(-2)=\frac{-15}{2}$

Now that I finish, I realise that you wanted f'(2); did you make a typo and actually need f'(-2), or did I just waste time? $:-\$

Yea you are right, I made a typo. Really sorry about that!

ashoni · « **Reply #340 on:** February 26, 2012, 04:13:51 pm »

Monique has calculated that the time taken,in minutes, for her to drive to work is 60-8m+m^2, where m is the number of minutes after 7 am that she leaves home.
a) How long does the trip tale of she leaves on time, at 7 am?
b) At what time can she leave so that the trip takes 44 minutes?
c) The journey will take 90 minutes if she leaves at what time?

stuck on this questions guys.. need help D:

Phy124 · « **Reply #341 on:** February 26, 2012, 04:47:15 pm »

Quote from: tony_123 on February 26, 2012, 04:13:51 pm

Monique has calculated that the time taken,in minutes, for her to drive to work is 60-8m+m^2, where m is the number of minutes after 7 am that she leaves home.
a) How long does the trip tale of she leaves on time, at 7 am?
b) At what time can she leave so that the trip takes 44 minutes?
c) The journey will take 90 minutes if she leaves at what time?

stuck on this questions guys.. need help D:

Firstly I'm just going to rearrange the equation to $f(m) = m^{2}-8m+60$ because I find it easier that way

a) 7am is 0 minutes after 7am so:

$f(0) = (0)^{2} - 8(0) + 60 = 60 minutes$

b) We want the result answer to equal 44 minutes so:

$m^{2} - 8m + 60 = 44$

$\therefore m^{2}-8m+16=0$

$(m-4)^{2}=0$

$\therefore m = 4$

This means it is 4 minutes after 7am so the time she can leave will be 7:04 am

c) We want the resultant answer to equal 90 minutes so:

$m^{2} - 8m + 60 = 90$

$\therefore m^{2} - 8m - 30 = 0$

$a=1, b=-2, c= -30$

$m = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$m = \frac{-(-8) \pm \sqrt{(-8)^{2}-4(1)(-30)}}{2(1)}$

$m = \frac{8 \pm \sqrt{184}}{2}$

$m = \frac{8}{2} \pm \frac{\sqrt{184}}{2}$

$m = 4 \pm \frac{\sqrt{46 \times 4}}{2}$

$m = 4 \pm \frac{2 \sqrt{46}}{2}$

$m = 4 \pm \sqrt{46}$

This means our answer for m will be:

$m = 4 + \sqrt{46}$

As the alternate answer will be negative and we are not leaving before 7am

Now add this answer to 7am (i.e. $7:(4 + \sqrt{46}) am$ - I don't have a calculator, but it would probably be ~7.10-7.11 am

ashoni · « **Reply #342 on:** February 26, 2012, 04:55:15 pm »

thanks so much Phy124! you make everything look so clear and simple haha!

do you mind doing explaining this to me as well?
If 2x^2+p=3x:
a)find a value of p so that there are two rational solutions.
b)find another value of p that gives two rational solutions
c) find the solution to the equation for the values of p found in parts a and b.

thankss

paulsterio · « **Reply #343 on:** February 26, 2012, 05:31:27 pm »

That's basically a discriminant question.

$2x^2 -3x+p=0$

$\Delta = (-3)^2 - 4(2)(p) = 9-8p$

For 2 rational solutions,
$\Delta = 9-8p = a^2, a \epsilon Z$

$p = \frac{9-a^2}{8}$

So you can sub in any two integer values for a in order to get an answer for part a) and part b)

Then once you have a value of p, you can just solve the equation like a normal quadratic (which I assume you know how to solve) for part c)

Insa · « **Reply #344 on:** February 27, 2012, 11:09:29 pm »

Hello,

This question should be a piece of cake for you guys. I'm getting confused about factorising this. $50(3x+2)^{2}-98$

Cheers

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