General Discussion => Other Discussions => Math and Science => Topic started by: Yeong on September 21, 2011, 07:59:46 pm
Title: Calculus help?
Post by: Yeong on September 21, 2011, 07:59:46 pm
Ok. I confronted some question in Calculus class. I guessed the answers right, but I don't know how to get the answer.. XP Here's the question: Find 2 lines that are both tangent of graph y=x^{2} and y=-x^{2}+6x-5. Ans: y=2x-1, y=4x-4 I really got these right by guessing but I need proving. help! XP
Title: Re: Calculus help?
Post by: chattahippie on September 21, 2011, 08:05:09 pm
You find two points that only touch the two equations once each (a tangent), then calculate the slope for each using the slope formula, then b.
y2-y1 -------- = m x2-x1
Both of those lines touch each equation only once, making them each tangent to both y = x^2 and y = -x^2 + 6x - 5
Hope that helps
Title: Re: Calculus help?
Post by: Yeong on September 21, 2011, 08:06:55 pm
But in that way can't you get more than 2 lines? The answer was only those two line, and no more.
Title: Re: Calculus help?
Post by: fb39ca4 on September 21, 2011, 08:09:56 pm
No, because there are two pairs of two points each that will create tangent lines. If you mix and match points between the pairs, you will get lines that are not tangent.
Title: Re: Calculus help?
Post by: chattahippie on September 21, 2011, 08:12:17 pm
You have to make sure that the points you chose only touch each x^2 equation once, and stay on the outside edge, because if they were on the inside they would intersect each equation in 2 places. That narrows down the selection a lot, especially when you are dealing with tangents that must be tangent to two functions/equations. I'm actually not in calculus, but I really like math and from looking intensely at a 96 x 64 graph of all four equations, it does seem that those two lines are the only lines that perfectly work.
Title: Re: Calculus help?
Post by: Yeong on September 21, 2011, 08:25:12 pm
That's how I got my answer too. XP But I need algebraic solution... (My teacher will ask us)
Title: Re: Calculus help?
Post by: Xeda112358 on September 21, 2011, 08:29:17 pm
I will try to work out a way, but I also notice that the second equation is -(x-1)(x-5) if that helps...
Title: Re: Calculus help?
Post by: Yeong on September 21, 2011, 08:30:03 pm
I tried that but that didn't help any.. XP setting both equations' derivation equal didn't worked either XP I don't know what to start with XP
Title: Re: Calculus help?
Post by: chattahippie on September 21, 2011, 08:55:26 pm
Well, I guess this can be tied to circles. Circles can only have 2 tangent lines if they aren't touching. Maybe the same rule applies for parabolas whose curves are in opposite directions?
Title: Re: Calculus help?
Post by: sqrt(Time) on September 21, 2011, 08:56:19 pm
I'm not sure this is the most elegant way of solving it, but it sure should work...
Call the place where it touches the first curve (a,a^2). You find the derivative at the point a, and so you can find the slope and y-intercept of the line - as a function of a. You should have something like m=f(a) intercept=g(a)
Now call the place where it touches the second curve (b,-b^2 + 6b -5). You can also find the derivative there, and find the slope and intercept as a function of b, too - something like m=h(b) and intercept=i(b).
So now you have f(a)=h(b) g(a)=i(b)
Solve these two equations for a and b, then plug either of one of those into your functions to get the slope and intercept of the lines.
Spoiler For Spoiler:
f(a) and h(b) should be linear, and g(a) and i(b) should be quadratic.
Title: Re: Calculus help?
Post by: phenomist on September 23, 2011, 01:20:47 am
Another method might be to note that both parabolas have the same curvature, only negated. Their vertex points are at (0,0) and (3,4). Hence, any line tangent to both parabolas must lie on the midpoint of their vertices (1.5,2), which can be easily shown by rotating the diagram.
Considering the tangent line at (a,a^2) has a slope of 2a, we can equate slopes: (a^2-2)/(a-1.5)=2a => a^2-2 = 2a^2-3a => a^2-3a+2=0 => a=1 or 2.
These generate the desired lines, of slopes 2 and 4.