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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Mellyboo on June 01, 2015, 01:39:16 pm

Title: Heading due east
Post by: Mellyboo on June 01, 2015, 01:39:16 pm: It seems like a simple question... But im lost :o the question asks me to (the curve starts at the starting point (0,0) for all questions)" determine the parabolic path so that it is heading due east when it reaches marker D(12,-4)" it asks the same question for for a cubic function but finishing at point A(12,4) passing through P(6,-1) then again for a logarithmic question(without point P) finishing at A(12,4) but adds, why is it not possible to head due east in a logarithmic function? What is heading due east???!!
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 03:07:06 pm: Would really appreicate if someone could help , sac is tomorrow
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 03:35:54 pm: The gradient is 0 at those points
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 04:08:22 pm: Ah yeah I figured it out thanks . So how do I ensure that the parabola and cubic graph have turning points at the markers they are asking for ? I know I could use turning point form for the parabola , but what about the cubic ?
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 04:20:08 pm: Quote from: Mellyboo on June 01, 2015, 04:08:22 pm
Ah yeah I figured it out thanks . So how do I ensure that the parabola and cubic graph have turning points at the markers they are asking for ? I know I could use turning point form for the parabola , but what about the cubic ?
Differentiate the cubic, and let the derivative equal zero. You want the gradient to be 0 when x=12. So for a general cubic, f(x)=ax^3+bx^2+cx+d => f'(x)=3ax^2+2bx+c
=>0=3a(12)^2+2b(12)+c <---------(1)
You've got three other points so you can solve for a,b,c,d.
Title: Re: Heading due east
Post by: Special At Specialist on June 01, 2015, 04:45:10 pm: Quote from: grannysmith on June 01, 2015, 04:20:08 pm
Differentiate the cubic, and let the derivative equal zero. You want the gradient to be 0 when x=12. So for a general cubic, f(x)=ax^3+bx^2+cx+d => f'(x)=3ax^2+2bx+c
=>0=3a(12)^2+2b(12)+c <---------(1)
You've got three other points so you can solve for a,b,c,d.

Exactly what grannysmith said. You let the derivative be equal to zero.

To elaborate further:
For the quadratic, you have:
y = ax^2 + bx + c
y' = 2ax + b
To head due east, let y' = 0
2ax + b = 0
x = -b/(2a)
Substitute this in to the original equation to find the y value:
y = a*(-b/(2a))^2 + b*(-b/(2a)) + c
y = b^2 / (4a) - b^2 / (2a) + c
y = -b^2 / (4a) + c
y = (4ac - b^2) / (4a)
Given the point (12, -4), we can solve for a, b and c:
When x = 12, y = -4 and y' = 0:
12 = -b/(2a)
b/a = -24
b = -24a
-4 = (4ac - b^2) / (4a)
4ac - b^2 = -16a
ac - (b/2)^2 = -4a
Substitute b = -24a:
ac - 144a^2 = -4a
ac = 140a^2
a*(140a - c) = 0
a = 0 or 140a = c
We are assuming that a is not equal to zero, since it's a quadratic, not a linear equation.
Therefore 140a = c
So we have b = -24a and c = 140a
Now we use our original equation and the point (12, -4):
y = ax^2 + bx + c
y = ax^2 - 24ax + 140a
-4 = 144a - 288a + 140a
-4 = -4a
a = 1
b = -24(1) = -24
c = 140(1) = 140
So the equation is:
y = x^2 - 24x + 140

For the cubic, you do the same thing essentially, just with an extra equation and an extra variable.

For the logarithm question, you know that logarithms have a horizontal asymptote. That means that the curve will never end up horizontal, it will only approach zero gradient, so it will never head due east.
We can prove this algebraically as well:
Let y = a*ln(bx + c) + d
Try to solve for y' = 0:
ab / (bx + c) = 0
0*(bx + c) = ab
0*bx = ab - c
0x = (ab - c) / b
No value of x will turn 0x into (ab - c) / b
Therefore, this equation is inconsistent.
Therefore, the logarithmic equation will never have zero gradient.
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 06:09:18 pm: Cannot be x^2-24x+140.
You must have done something wrong
It has to start at 0 and have a turning point at (12,-4)
Solving i got 1/36(x-12)^2-4 which i verified by subbing the points back in.
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 06:12:16 pm: Quote from: grannysmith on June 01, 2015, 04:20:08 pm
Differentiate the cubic, and let the derivative equal zero. You want the gradient to be 0 when x=12. So for a general cubic, f(x)=ax^3+bx^2+cx+d => f'(x)=3ax^2+2bx+c
=>0=3a(12)^2+2b(12)+c <---------(1)
You've got three other points so you can solve for a,b,c,d.

If the derivative =0 for each of those points then they must be stationary points. How can we assume that each of the points given are stationary? so for the cubic it has to pass through (0,0), (6,-1) and have a stationary point at (12,4)... How would i solve this?
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 06:48:55 pm: If i understood the question correctly its going to look something like this,
http://imgur.com/7m7t893

I just have no idea how to work out the equation knowing that one has to be a turning point
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 07:09:18 pm: Quote from: Mellyboo on June 01, 2015, 06:12:16 pm

If the derivative =0 for each of those points then they must be stationary points. How can we assume that each of the points given are stationary? so for the cubic it has to pass through (0,0), (6,-1) and have a stationary point at (12,4)... How would i solve this?
We're manipulating the equation so that there is a stationary point at (12,4). The other points (e.g. (0,0), (6,-1)) are just coordinates that the cubic passes through - they're not stationary points.

So let's summarise all the information we have about this cubic:
-it passes through (6,-1)
-it passes through (0,0)
-it also passes through (12,4)
-but (12,4) also happens to be a stationary point

Do you see how we can set-up 4 different equations to solve for the four unknowns? Like so:

$\text{Let} y=ax^3+bx^2+cx+d$
$\implies \frac{dy}{dx}=3ax^2+2bx+c$
$\text{We want it to have a stationary point at (12,4). So at x=12,} \frac{dy}{dx}=0$
$\implies 0=3a(12)^2+2b(12)+c \text{This is your first equation. We need another 3.}$
$\text{When x=0, y=0:} \rightarrow 0=d$
$\text{When x=6, y=-1:} \rightarrow -1=a(6)^3+b(6)^2+c(6)+0$
$\text{When x=12, y=14:} \rightarrow 14=a(12)^3+b(12)^2+c(12)+0$
Now you can solve for the unknowns
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 08:18:48 pm: Assuming you meant y= 4 for the last equation yes?
Anyways thank you so much!!!!!
Last question, if they ask me to further dilate a parabolic equation (actually its a hybrid consisting of two parabolas, one negative and one positive) by 1.3 do i just multiply 1.3 with the a value for both equations? Both equations I worked out so thay the gradient is the same when they join ( smooth) but once i multiplied the a values by 1.3 , they didnt touch anymore... Im assuming that they still need to join or is this not possible?
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 08:23:42 pm: http://imgur.com/j52N9SL

Ahhhh it came up as false! What am i doing wrong??
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 08:26:02 pm: Quote from: Mellyboo on June 01, 2015, 08:18:48 pm
Assuming you meant y= 4 for the last equation yes?
Anyways thank you so much!!!!!
Last question, if they ask me to further dilate a parabolic equation (actually its a hybrid consisting of two parabolas, one negative and one positive) by 1.3 do i just multiply 1.3 with the a value for both equations? Both equations I worked out so thay the gradient is the same when they join ( smooth) but once i multiplied the a values by 1.3 , they didnt touch anymore... Im assuming that they still need to join or is this not possible?
Ah yes, y=4. My bad.

You have to multiply the entire function by 1.3. E.g. if f(x)=x^2+x+1, then dilating by a factor of 1.3 from the x-axis --> 1.3*f(x)=1.3*(x^2+x+1)
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 08:28:21 pm: Quote from: Mellyboo on June 01, 2015, 08:23:42 pm
http://imgur.com/j52N9SL

Ahhhh it came up as false! What am i doing wrong??
Remember that you're not cubing the 'a' or squaring the 'b' - you're only cubing/squaring their coefficients. ax^2 = a*x^2 and does NOT equal (ax)^2
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 08:34:17 pm: Yeah the calculator did that by default. I tried fixing it and still false. http://imgur.com/iPxKCaF
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 08:39:56 pm: And thank you for the dilation one, it worked :) cubic still not though:(
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 08:55:21 pm: Quote from: Mellyboo on June 01, 2015, 08:34:17 pm
Yeah the calculator did that by default. I tried fixing it and still false. http://imgur.com/iPxKCaF
Not sure what's happening there - I'm getting a=-1/54, b=5/12 and c=-2. Perhaps try multiplying each term out instead of using brackets?
Title: Re: Heading due east
Post by: Mellyboo on June 01, 2015, 09:11:27 pm: Yep that did it. Thank you thank you thank you !!!!!!!!!!! ;D also is the way 'special at specialist' explained why the log cannot have a " due east" correct?
Title: Re: Heading due east
Post by: grannysmith on June 01, 2015, 09:14:41 pm: Quote from: Mellyboo on June 01, 2015, 09:11:27 pm
Yep that did it. Thank you thank you thank you !!!!!!!!!!! ;D also is the way 'special at specialist' explained why the log cannot have a " due east" correct?
Happy to help :)

Yes, that's the correct method.
Title: Re: Heading due east
Post by: Sundal on June 02, 2015, 08:01:25 pm: Not to barge in, but I didn't know log graphs had a horizontal asymptote?

I'm a little confused, can someone explain?

Cheers.
Title: Re: Heading due east
Post by: Phy124 on June 03, 2015, 03:13:07 am: Quote from: Sundal on June 02, 2015, 08:01:25 pm
Not to barge in, but I didn't know log graphs had a horizontal asymptote?

I'm a little confused, can someone explain?

Cheers.
They don't, it was a poor choice of words on Special At Specialist's part, however he was correct in saying "that the curve will never end up horizontal, it will only approach zero gradient, so it will never head due east."

It is correct that if a function exhibits the behaviour of a horizontal asymptote e.g. as $x \rightarrow \infty, y \rightarrow a$ for some $a\in \mathbb{R}$ that this implies as $x \rightarrow \infty, \frac{dy}{dx} \rightarrow 0$ . However it is not true the other way around i.e. As $x \rightarrow \infty, \frac{dy}{dx} \rightarrow 0$ then $x \rightarrow \infty, y \rightarrow a$ for some $a\in \mathbb{R}$ .

Take $y = \log_e(x)$ for example, where $\frac{dy}{dx}=\frac{1}{x}$ .

Based on the above we can say that as $x \rightarrow \infty, \frac{dy}{dx} \rightarrow 0$ .

However, if the false implication was true then we would say $\log_e(x)$ has a horizontal asymptote. Why have we never drawn one in during our years of logarithmic study then? Well $\log_e(x)$ is strictly increasing for all $x>0$ , so it can't have a horizontal asymptote.

Feel free to read this post and its accompanying comments for some discussion/proofs/proper explanations on why the logarithm function doesn't have a horizontal asymptote.
Title: Re: Heading due east
Post by: Sundal on June 03, 2015, 06:00:37 pm: Woah, thanks for the in-depth explanation Phy124! ! It was truly wonderful!