quadratics

cooljackiec · 2014-04-01 12:59:40

What is the maximum value of c such that the graph of the parabola

has at most one point of intersection with the line x+c?

Consider the ellipse

Find the maximum value of the product xy on the ellipse.

Agnishom · 2014-04-01 13:22:39

Hint:

For question 1, We need the intersection of the two graphs y = 1/3 x^2 and y = x + c. Subtract one eqaution from the other to get 1/3 x^2 - x - c = 0. Since there are only one solutions to this quadratic equation, the discriminant must be 0. Equate the discriminant with 0 and solve for c. If c is a constant you are done. If c is a function of x, maximise the function by equating the derivative of that function with 0.

P.S : I am not at home. So, I do not have a powerful CAS or pen and paper to work with now.

bobbym · 2014-04-01 13:32:54

Hi;

Agnishom · 2014-04-01 13:39:12

For question 2,

Solve the equation of the ellipse for y to get y as a function of x

x^2 + 2y^2 = 16
or 2y^2 = 16 - x^2 = (4-x) (4+x)

Multiply both sides with x,

Let us call the right side f(x)

Calculate the derivative of the right side. Let that derivative be f'(x)
Equate f'(x) with 0
Solve for x

If you get one value of x, skip the next paragraph.

If you get more than one values of x, check on the second derivative. If the second derivative in that point is positive, the corresponding value you are getting is not the minimum. Continue checking other values until you are one a point where the second derivative is negative. Select this value.

Compute f(x). That is the answer.

[I can't compute that derivative by hand, so no answers now]

Agnishom · 2014-04-01 13:44:06

Hi bobbym,

Is my approach on the problem correct?

By the way, I am attending a funeral

ShivamS · 2014-04-01 13:48:03

You get two answers for x.

Sorry for your loss.
You are going to a funeral this early? I think it's only around 7:15 AM there.

Last edited by ShivamS (2014-04-01 13:51:10)

bobbym · 2014-04-01 13:50:14

Hi Agnishom;

That is what I am getting using another approach. Sorry about your loss.

Is my approach on the problem correct?

I got most of your method using a geometric approach.

Agnishom · 2014-04-01 14:15:13

How can I calculate that derivative by hand. I had to go to a distant village, so I am there this early.

bobbym · 2014-04-01 14:19:21

You would use the standard methods for differentiation. Or you might try a different method...

Agnishom · 2014-04-01 14:23:02

Can the other method work without a computer? If no, then you'll have to wait until I get back home

bobbym · 2014-04-01 14:25:20

You are obviously on the internet, take it over to Wolfram if you can not do it with pencil and paper.

There is a book called Maxima and Minima Without Calculus by Niven. Hunt it down you might like it.

Also, you can get a decent approximation to the derivative using differences. Want to see how?

eigenguy · 2014-04-01 15:40:47

Agnishom wrote:

For question 1, We need the intersection of the two graphs y = 1/3 x^2 and y = x + c. Subtract one eqaution from the other to get 1/3 x^2 - x - c = 0. Since there are only one solutions to this quadratic equation, the discriminant must be 0. Equate the discriminant with 0 and solve for c. If c is a constant you are done. If c is a function of x, maximise the function by equating the derivative of that function with 0.

This starts off well, but goes astray. If we are looking for an intersection point then the same value of x gives the same y in both equations, so 1/3 x^2 = x + c, or 1/3 x^2 - x - c = 0.
At this point, you pull out your handy-dandy quadratic equation to get x = (something) +/- (something) √((-1)^ - 4(1/3)(-c)) = (something) +/- (something)√(1 - (4/3)c), where I've not bothered with the somethings, because we actually don't care about the value of x. What we need, though, is for there to be 0 or 1 real values for x, this occurs only when the discrimant (1-(4/3)c) <= 0. Thus bobbym's solution.

c is never a function of x, here.

As for question 2, one approach (whether it was bobbym's or not, I don't know), is to consider the curves xy = k. These are a family of nonoverlapping hyperboli with the axes as asymptotes. As the value of k increases, the hyperboli move farther away from the origin. So consider the intersection of the ellipse with the hyperbola xy = k for a particular k. For small enough k, this intersects with ellipse in 2 points in quadrant I (and 2 more in Q3, but they are mirror images, and give nothing new). As we increase k, the hyperbola moves outward, and the intersection points move closer together. Increase k far enough, and the intersection points meet, leaving a single point of tangency. This is the maximum value of k for which the hyperbola intersects the ellipse. Increase k any more and there is no intersection. That k cannot be reached on the ellipse.

So the maximum value k of xy occurs when the hyperbola xy = k is tangent to the ellipse. As with question 1, substitute y = k/x into the ellipse equation, solve for x^2, find the discriminant. The value of k that makes the discrimant = 0 is the maximum value of xy on the ellipse.

Last edited by eigenguy (2014-04-01 15:57:03)

Agnishom · 2014-04-01 16:36:16

Looks like I am totally out of track...

bobbym · 2014-04-01 20:05:52

What did you get for an answer?

cooljackiec · 2014-04-02 10:37:23

well for agnishom's first one, with the tangent to parabola, I got two values so naturally i took the smallest one. I somewhat get the ellipse solution

bobbym · 2014-04-02 14:06:05

Hi cooljackiec;

Sorry, I was asking Agnishom.

Agnishom · 2014-04-03 18:30:23

Hi bobbym;

How do I find the derivative of something in M? The manual just says '

bobbym · 2014-04-03 20:28:15

That is one way.

D[x Sqrt[((4 - x) (4 + x))/2], x] // FullSimplify

Agnishom · 2014-04-03 20:29:43

Thank You.

How do you do this with Geogebra?

bobbym · 2014-04-03 20:33:10

Geogebra uses maxima to do that.

Agnishom · 2014-04-03 20:35:38

I've an Old Java Version of Geogebra. Does the latest Geogebra have Maxima in it?

bobbym · 2014-04-03 20:39:42

You have to get the one with the CAS. It is larger, approximately 65 MB.

I will see if just geogebra can do it.

bobbym · 2014-04-03 20:56:22

derivative[x*sqrt((4-x)(4+x)/2),x] is how in geogebra.

Agnishom · 2014-04-03 21:07:57

Will it give a picture?

bobbym · 2014-04-03 21:14:57

Yes, it will graph the derivative. For some reason your answer is 1 / 2 the actual answer.

Math Is Fun Forum

#1 2014-04-01 12:59:40

quadratics

#2 2014-04-01 13:22:39

Re: quadratics

#3 2014-04-01 13:32:54

Re: quadratics

#4 2014-04-01 13:39:12

Re: quadratics

#5 2014-04-01 13:44:06

Re: quadratics

#6 2014-04-01 13:48:03

Re: quadratics

#7 2014-04-01 13:50:14

Re: quadratics

#8 2014-04-01 14:15:13

Re: quadratics

#9 2014-04-01 14:19:21

Re: quadratics

#10 2014-04-01 14:23:02

Re: quadratics

#11 2014-04-01 14:25:20

Re: quadratics

#12 2014-04-01 15:40:47

Re: quadratics

#13 2014-04-01 16:36:16

Re: quadratics

#14 2014-04-01 20:05:52

Re: quadratics

#15 2014-04-02 10:37:23

Re: quadratics

#16 2014-04-02 14:06:05

Re: quadratics

#17 2014-04-03 18:30:23

Re: quadratics

#18 2014-04-03 20:28:15

Re: quadratics

#19 2014-04-03 20:29:43

Re: quadratics

#20 2014-04-03 20:33:10

Re: quadratics

#21 2014-04-03 20:35:38

Re: quadratics

#22 2014-04-03 20:39:42

Re: quadratics

#23 2014-04-03 20:56:22

Re: quadratics

#24 2014-04-03 21:07:57

Re: quadratics

#25 2014-04-03 21:14:57

Re: quadratics

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