Parametric Parabolas

It hasn't even been a week since we dealt with parabolas, and they come again, this time with actual numbers. The question itself isn't hard, but once again, some thinking is required.

A parabola goes through the points A(17, 3) B(34, z) C(50, -30). B is the maximum point of the parabola. Find point B and the parabola function in standard form.

As you know, a parabola can be written in two ways. One is standard form, with the a, b, and c coefficients, and one is the vertex form. To solve this question, we'll use the vertex form. This form says that the equation a parabola that has a vertex (h, k) is:
y = a(x - h)² + k

We know h and k, so let's plug them in:
y = a(x - 34)² + z

We have two other points, so let's use them:
3 = a(17 - 34)² + z
2 = a(50 - 34)² + z

3 = 289a + z
-30 = 256a + z

Subtract the second equation from the first:
33 = 33a
a = 1
3 = 289a + z
3 = 289 + z
z = -286

The function of the parabola is:
y = (x - 34)² - 286

Open parenthese and combine like terms to switch to standard form:
y = x² - 68x + 1156 - 286
y = x² - 68x + 870

Nice, isn't it?
Nadav

nadavs

Inverse Quadratic Function

Today I have another question in the genre of "short question, long answer". Today's question deals with inverse functions, more accurately the inverse of a quadratic function (or more accurately, what becomes a quadratic function). This is not some simple quadratic function, but rather something more difficult to solve.

Find the inverse of y = 2x + sqrt(x)

Short, isn't it? However, the solution isn't that short.

To find the inverse, we need to substitute x and y and then solve for y. Let's do it:
x = 2y + sqrt(y)

To eliminate the square root, move the 2y to the right side and square:
x - 2y = sqrt(y)
x² - 4xy + 4y² = y

Subtract y from both sides and use the distributive property:
x² - 4xy - y + 4y² = 0
x² - y(4x + 1) + 4y² = 0

This calls for the friend of any math student: the quadratic formula. Notice that squaring may add a solution, but there is only one inverse function, so we'll need to eliminate one:
y = ((4x + 1) ± sqrt((4x + 1)² - 16x²)) / 8

Yes, not a pleasant look. Let's work with that a little so see if it gets any better:
y = ((4x + 1) ± sqrt(16x² + 8x + 1 - 16x²)) / 8
y = ((4x + 1) ± sqrt(8x + 1)) / 8

That's the best it gets, but there's still a problem: there are two possible inverse functions here, but only one is right. To find out which one is right, remember that we reached the following conclusion:
x - 2y = sqrt(y)

This means x - 2y must be non-negative. Let's plug the two possible y values to see what we get:
x - ((4x + 1) + sqrt(8x + 1)) / 4 = 4x / 4 - ((4x + 1) + sqrt(8x + 1)) / 4
= (4x - ((4x + 1) + sqrt(8x + 1))) / 4 = (-1 - sqrt(8x + 1)) / 4
In one word: negative. Not good.

Let's plug the other one now to see if it works:
x - ((4x + 1) - sqrt(8x + 1)) / 4 = 4x / 4 - ((4x + 1) - sqrt(8x + 1)) / 4
= (4x - ((4x + 1) - sqrt(8x + 1))) / 4 = (-1 + sqrt(8x + 1)) / 4

As you can clearly see, sqrt(8x + 1) > 0 if x > 0, just like the domain of the original function. That means the inverse of the original function is:
y = ((4x + 1) - sqrt(8x + 1)) / 8

Here is a little confirmation test:
Plug 9 for x in the original function. You get y = 2*9 + sqrt(9) = 21. Now plug 21 in the inverse function and see if it gives back 9:
y = ((4 * 21 + 1) - sqrt(8 * 21 + 1)) / 8
y = (85 - sqrt(168 + 1)) / 8
y = (85 - sqrt(169)) / 8
y = (85 - 13) / 8
y = 72 / 8 = 9

Yes, it is the inverse.

You can find many more interesting math tricks like that on super math tips.

Hope you liked it.
Nadav

nadavs

Cubic Funcion Discovery

Today I have another question from Yahoo Answers (super math tips people, send some questions too!). The question is simple, but its answer is very, very long.

There is a cubic function that passes through the following points: (-2, -24) (3, -4) (5, 18) (6, 56). Find out what the function is.

(The original question had a point different than (6,56), but it makes it shorter, so the fun is gone).

At first, it seems impossible. But then, you remember that a cubic function has the following form:
y = ax³ + bx² + cx + d

We have x and y, so we can plug them in to find a, b, c, and d! All we have to do now is set up a system of four equations and solve them. All of them (here comes the long part):

I -8a + 4b - 2c + d = -24
II 27a + 9b + 3c + d = -4
III 125a + 25b + 5c + d = 18
IV 216a + 36b + 6c + d = 56

Now we need to get three equations out of those four. We do that be subtracting the equations from each other and getting rid of d. If done smart, it can even simplify the process in the future (if we can get small enough coefficient). There are six possible outcomes, and the smartest thing to do is to choose the best ones:

A. 91a + 11b + c = 38 (IV - III)
B. 98a + 16b + 2c = 22 (III - II)
C. 35a + 5b + 5c = 20 (II - I)

Now we need to eliminate another variable and create two equations. The chosen one is c. First, we multiply equation A by 2:
A2: 182a + 22b + 2c = 76
B2: 98a + 16b + 2c = 22

After we subtract the two equations we get:
84a + 6b = 54
Divide by 6 to simplify:
14a + b = 9

Now multiply equation A by 5:
A5: 455a + 55b + 5c = 190
C5: 35a + 5b + 5c = 20

Subtract them to get:
420a + 50b = 170
Divide by 10:
42a + 5b = 17

Now we have two more equations:
i. 42a + 5b = 17
ii. 14a + b = 9

Multiply equation ii by 5:
70a + 5b = 45

So now we have:
i21: 42a + 5b = 17
ii21: 70a + 5b = 45

Subtract the two equations:
28a = 28
a = 1

Now everything is going to be super fast. We just need to plug in the numbers we get, this time in equation ii (14a + b = 9):
14 * 1 + b = 9
14 + b = 9
b = -5

Another coefficient! They go down like flies! Now let's use equation A, since it has c with a coefficient of 1:
91a + 11b + c = 38
91 - 55 + c = 38
c = 2

So tiny, with so many problems. One coefficient to go, d, the y-intercept. Let's plug it in equation II, where all coefficients are positive but small:
27a + 9b + 3c + d = -4
27 - 45 + 6 + d = -4
d - 12 = -4
d = 8

THE END!

We now have all coefficients for this cubic function, and we can proudly say that it is:
y = x³ - 5x² + 2x + 8

Who said cubics are hard?

Nadav


nadavs

Buggy Calculus

Today I have a question of a higher level from the previous two, but it's much easier from the previous ones for people who know calculus (seriously, much easier).

So, here is the question:

A bug is walking on a wooden log. Its distance from the right edge of the log after t minutes is given by the function x(t) = t³ - 9t² + 500 inches.

A. What is the bug's velocity after 4 minutes?
B. When does the bug change direction?
C. When does the bug's acceleration equal zero?

If you don't know calculus you'd probably say "There's no way to know that", but there is a way, and it's simple.

The velocity function of the bug is the derivative of its distance function. The acceleration of the bug is the derivative of its velocity function. When we get those two functions, the question is very easy.

A. Let's find v(t), the bug's velocity function:
x'(t) = v(t) = 3t² - 18t .
To find the velocity after four minutes, we just plug in 4 instead of t:
4² - 18 * 4 = 16 - 72 = -56 inches/minute. A negative velocity means that the bug is going "backwards", meaning towards the right edge (because we defined a positive distance as going away from the right edge, meaning left).

B. To find when the bug changes direction, we need to find out when its velocity is zero. That means the velocity is changing signs, and by changing signs it means a change of direction. We already have the function, so let's make it equal zero:
3t2 - 18t = 0
t2 - 6t = 0
t(t - 6) = 0
t = 0, 6

The bug cannot change direction at minute zero, since it did not have a direction before. That means the bug changes direction on the 6th minute.

C. To find when the acceleration is zero, we first have to find the acceleration function. We do that by derivating the velocity function:
v'(t) = a(t) = 6t - 18.
Now let's make it equal to zero:
6t - 18 = 0
6t = 18
t = 3

That means the bug's acceleration is zero after three minutes.

Don't understand a word of this post? Don't be shy to ask anything on this blog's comments or the super math tips.

Hope you learned something.
Yours,
Nadav

nadavs