Author Topic: VCE Specialist 3/4 Question Thread! (Read 2827983 times) Tweet Share

polar · « **Reply #1500 on:** March 26, 2013, 06:59:47 pm »

Quote from: Homer on March 26, 2013, 06:18:17 pm

dont know how to

$\begin{aligned} \frac{d}{dx}\left(x^3 + \frac{2x}{y}\right) &= \frac{d}{dx}(3) \\ 3x^2 + \frac{2\times y - 2x\times\frac{dy}{dx}}{y^2} = 0 \\ \frac{2\times y - 2x\frac{dy}{dx}}{y^2} = -3x^2 \\ 2y-2x\frac{dy}{dx} = -3x^2y^2 \\ \frac{dy}{dx} = \frac{2y+3x^2y^2}{2x} \end{aligned}$

Substitute x=1,y=2 into that and then use $y-y_1=m(x-x_1)$

Stick · « **Reply #1501 on:** March 26, 2013, 07:02:21 pm »

On the topic of implicit differentiation, I'm stuck with part c of the question below (I managed to skip it and work out part d OK).

b^3 · « **Reply #1502 on:** March 26, 2013, 08:41:27 pm »

Misread what post was asking for..., refer to Polar's post below.

$\begin{alignedat}{1}x^{3}+xy+2y^{3} & =k \\ \frac{d}{dx}\left(x^{3}\right)+\frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(2y^{3}\right) & =\frac{d}{dx}\left(k\right) \\ 3x^{2}+x\frac{dy}{dx}+y\left(1\right)+6y^{2}\frac{dy}{dx} & =0 \\ \frac{dy}{dx}\left(x+6y^{2}\right) & =-3x^{2}-y \\ \frac{dy}{dx} & =\frac{-3x^{2}-y}{x+6y^{2}} \end{alignedat}$

For the tangent to be vertical, $\frac{dx}{dy}=0.$
$ \begin{alignedat}{1}\frac{x+6y^{2}}{-3x^{2}-y} & =0 \\ x & =-6y^{2} \end{alignedat} $

Substitute this back into the equation for the curve.
$\begin{alignedat}{1}\left(-6y^{2}\right)^{3}+\left(-6y^{2}\right)y+2y^{3} & =k \\ -216y^{6}-6y^{3}+2y^{3} & =k \\ 216y^{6}+4y^{3}+k & =0 \end{alignedat}$

polar · « **Reply #1503 on:** March 26, 2013, 08:41:45 pm »

Quote from: Stick on March 26, 2013, 07:02:21 pm

On the topic of implicit differentiation, I'm stuck with part c of the question below (I managed to skip it and work out part d OK).

(Image removed from quote.)

when you solve $216y^6 + 4y^3 + k=0$ , you must get a solution otherwise there wouldn't be a tangent parallel to the y-axis.

$216y^6 + 4y^3 +k = 216(y^3)^2 + 4y^3 +k =0$ , for there to be a solution, $4^2-4(216)k \ge 0 \implies k \le \frac{1}{54}$ (discriminant has to be greater or equal to 0)

b^3 · « **Reply #1504 on:** March 26, 2013, 08:45:06 pm »

Ok I somehow read part b instead of part c.... well the working is there for part b anyways.. if anyone else wants it.... I really need sleep...

Stick · « **Reply #1505 on:** March 26, 2013, 09:05:55 pm »

Quote from: polar on March 26, 2013, 08:41:45 pm

when you solve $216y^6 + 4y^3 + k=0$ , you must get a solution otherwise there wouldn't be a tangent parallel to the y-axis.

$216y^6 + 4y^3 +k = 216(y^3)^2 + 4y^3 +k =0$ , for there to be a solution, $4^2-4(216)k \ge 0 \implies k \le \frac{1}{54}$ (discriminant has to be greater or equal to 0)

Ah, I didn't realise it was wanting me to use the discriminant. Thanks, polar and b^3.

Holmes · « **Reply #1506 on:** March 26, 2013, 09:22:22 pm »

Quote from: Holmes on March 26, 2013, 05:32:17 pm

I'm really confused with sketching rational functions. Is there a good way to sketch them other then using addition of ordinates. I know how to do reciprocal function sketching, but sometimes it's really hard to use that.

For example, I'm having trouble especially with this one:

$x$ $/$ $x^2 +1$

My teacher said that we can look at what happens at x approaches +/- infinity, but I'm still really confused.

Is this to go unanswered?

lzxnl · « **Reply #1507 on:** March 26, 2013, 10:45:06 pm »

You would have to firstly see what happens when x becomes really large. When x is large, the x^2 in the denominator is much greater than the x term in the numerator, so as x=>inf, x/(x^2+1)=>0
You can work out from which direction by considering the signs.
Now for vertical asymptotes. There are none as the denominator is never zero and the function is well-behaved.
We need to find stationary points of this function. Differentiating x/(x^2+1) yields (x^2+1-2x^2)/((x^2+1)^2)=(1-x^2)((x^2+1)^2)
Obviously we have stationary points when x=+-1
You can verify that x=1 is a local maximum and x=-1 is a local minimum by realizing that the function also has to pass through the origin.
So...x axis is an asymptote, passes through origin, coordinates of stationary points have been found, behaviour as x becomes large has also been determined...final thing is that the function is odd so you want to keep it symmetrical. Other than that, that's really it for sketching this.

zvezda · « **Reply #1508 on:** March 27, 2013, 06:34:02 pm »

Hey,
just wondering how to go about finding the derivative of:
dy/dx=(-x-y)/(x+3y)

Holmes · « **Reply #1509 on:** March 27, 2013, 06:49:07 pm »

Quote from: nliu1995 on March 26, 2013, 10:45:06 pm

the function is well-behaved.

Thanks a lot for your explanation!

zvezda · « **Reply #1510 on:** March 27, 2013, 06:49:29 pm »

Quote from: Stick on March 26, 2013, 09:05:55 pm

Ah, I didn't realise it was wanting me to use the discriminant. Thanks, polar and b^3.

how come you have used the formula for the discriminant? I thought that this only applied for quadtratics?

Lasercookie · « **Reply #1511 on:** March 27, 2013, 06:58:26 pm »

Quote from: stankovic123 on March 27, 2013, 06:49:29 pm

how come you have used the formula for the discriminant? I thought that this only applied for quadtratics?

Take a close look at the equation and how he's factored it. He's pretty much let y^3=u, which gives you a quadratic.

Henreezy · « **Reply #1512 on:** March 27, 2013, 07:34:31 pm »

I'm stuck on this question:

find dy/dx in terms of both x and y for the equation 3x^2+18x-y^2+4y+11=0,
then find the coordinates for which the tangent is parallel to the y axis.

e^1 · « **Reply #1513 on:** March 27, 2013, 07:47:48 pm »

Quote from: Henreezy on March 27, 2013, 07:34:31 pm

I'm stuck on this question:

find dy/dx in terms of both x and y for the equation 3x^2+18x-y^2+4y+11=0,
then find the coordinates for which the tangent is parallel to the y axis.

Spoiler

$\begin{align*} 3x^2+18x-y^2+4y+11=0 \\\\ \text{Using implicit differentiation:} \\\\ 6x + 18 - 2 \cdot \frac{dy}{dx} \cdot y + 4\frac{dy}{dx} &= 0 \\ 6x + 18 &= 2 \cdot \frac{dy}{dx} \cdot y - 4\frac{dy}{dx} \\ 6x + 18 &= \frac{dy}{dx}( 2y - 4) \\\\ \therefore \frac{dy}{dx} = \frac{6x + 18}{2y - 4} \end{align*}$

When the tangent's gradient is parallel to the y-axis, it is undefined. Hence, we should find when the denominator of $\frac{dy}{dx}$ equals 0:

$2y - 4 = 0 \implies y = 2$

Using the original equation:

$\begin{align*} 3x^2 + 18x - 4 + 8 + 11 &= 0 \\ 3x^2 + 18x + 15 &= 0 \\\\ \therefore x = -5 \text{ or } x= -1 \end{align*}$

Therefore at points (-5, 2) and (-1, 2) there is a tangent at these points which is parallel to the y-axis.

Also, check if I'm wrong. This I learned from some other calc book.
And about the undefined question: You can figure that out

polar · « **Reply #1514 on:** March 27, 2013, 07:49:57 pm »

Quote from: stankovic123 on March 27, 2013, 06:34:02 pm

Hey,
just wondering how to go about finding the derivative of:
dy/dx=(-x-y)/(x+3y)

$\begin{aligned} (x+3y)\frac{dy}{dx} &= -x-y \\ \text{Differentiate both sides (use product rule on left hand side)}: \\ \frac{dy}{dx}\times\frac{d}{dx}(x+3y) + (x+3y)\times\frac{d}{dx}\left(\frac{dy}{dx}\right) &= -1 - \frac{dy}{dx} \\ \frac{dy}{dx}\times\left(1+3\frac{dy}{dx}\right) + (x+3y)\frac{d^2y}{dx^2} &= -1-\frac{dy}{dx} \\ (x+3y)\frac{d^2y}{dx^2} + 3\left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx}+1 &=0\end{aligned}$

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2827983 times) Tweet Share

polar

Re: Specialist 3/4 Question Thread!

Stick

Re: Specialist 3/4 Question Thread!

b^3

Re: Specialist 3/4 Question Thread!

polar

Re: Specialist 3/4 Question Thread!

b^3

Re: Specialist 3/4 Question Thread!

Stick

Re: Specialist 3/4 Question Thread!

Holmes

Re: Specialist 3/4 Question Thread!

lzxnl

Re: Specialist 3/4 Question Thread!

zvezda

Re: Specialist 3/4 Question Thread!

Holmes

Re: Specialist 3/4 Question Thread!

zvezda

Re: Specialist 3/4 Question Thread!

Lasercookie

Re: Specialist 3/4 Question Thread!

Henreezy

Re: Specialist 3/4 Question Thread!

e^1

Re: Specialist 3/4 Question Thread!

polar

Re: Specialist 3/4 Question Thread!

Recent Posts

User:
Password: