Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2005-12-09 11:09:56

Registered: 2005-08-22
Posts: 1,504

parabolla's and velocity

I was solving an equation for velocity, by differentiating the equation of position. I had to find the values of the f(x) for which the velocity is zero, greater then zero, or less then zero. But the velocity equation was a quadratic. So the velocity was zero at two points. But the times when the velocity was greater or less then zero, thats not so simple. The graph of the quadratic will be a parabolla that oppens either upward or downard. It will cross the x axis at the zero's of the equation. Now if it opens upward, then the points between the two zero's will be below the x axis. If it opens downward, the points between the two zero's will be above the x axis.

We can find the region between the zero's by simply designating it to be the region less then the larg zero and greater then the small zero. And the region outside the zero's by the points that are either greater then the large or less then the small.

So all we need to do is find out whether the parabolla opens upward or downward. Thats easy, if the equation is ax^2 + bx + c, then if a is positive, it will be large and positve for large absolute values of x (opens upward). If a is negative, then it will be large and negative for large absolute values of x.

So now that we know whether it opens upward or downward, then we know where the region between the zero's is positive  or negative and if the region outside the zero's is positive or negative. So we now know the region where the velocity is positve or negative and have found the answer to the problem.

Is this cheating? Is there a more elegent way I could have solved this? A way to find the points on a parabolla that are above or below the x axis without considering its graph and whether the parabolla opens upward or downward?

The graph of the velocity in this problem was a parabolla which is pretty easy since it only has one vertice. But an equation of higher degree would be a nightmare! Is there any simple method to determine all the points above or below the graph of a function? You can't always say "f(x) > 0" or "f(x) < 0" and solve, because many functions have more then one zero and some of the roots will not be consitant with the inequality.

Another strange problem I got. Find the postive number that exceeds its cube by the greatest amount.

At first I tried differentiating c = x^3


then setting it equal to zero:

3x^2 = 0

Solved, x = 0. Thats obviously incorrect. I was unsure how to solve the problem and tried:

x > x^3


1 > 3x^2

1/3 > x^2        divided both sides by 3

sqrt(1/3) > x    square root of both sides

sqrt(3)/3 > x   rationalized denominator

This expression is an inequalit, but the answer to the problem was x equals sqrt(3)/3.

Peculiar. If I had set x = x^3, differentiated, and solved. That would have given me the answer.

But why? It doesn't seem like a normal maximums and minimums problem.

A logarithm is just a misspelled algorithm.


#2 2005-12-09 11:18:13

Registered: 2005-06-22
Posts: 4,900

Re: parabolla's and velocity

I don't think there is a simpler method. You just sketch the graph to see what it looks like, find the roots algebraically and combine the result(s) with the graph to find the regions. If it is a complicated function, you could use numerical method to estimate the roots to the required degree of accuracy instead of calculating them.

For your second question, you want to find the highest value of x - x³.

Differentiate: dy/dx = 1 - 3x²
Equate to 0: 1 - 3x² = 0
3x² = 1
x² = 1/3
x = 1/√3

Check to see if it is a maximum:
d²y/dx² = -6x. At x = 1/√3, d²y/dx² is negative, so it is a maximum point.

Edit: Oops, you said pretty much all of that already. I must need sleep. smile

Why did the vector cross the road?
It wanted to be normal.


#3 2005-12-09 13:49:38

Registered: 2005-08-22
Posts: 1,504

Re: parabolla's and velocity

No I didn't. I accidently got the answer by guessing. It resulted in an inequality giving me a half correct answer. The equation x^3 = x is the same as x - x^3, but when its in this form then it makes sense with maximums and minimums. x - x^3 will give us the difference and show us how much x exceeds its cube for any value of x. The high point of the graph is the value of x for which it exceeds it the most! :-D
Its pure genius! yikes

Last edited by mikau (2005-12-09 13:51:55)

A logarithm is just a misspelled algorithm.


Board footer

Powered by FluxBB