Finding a point on a slope that meets a function - tricky one though

[Femme_physics]

Homework Statement

Find the coordinates of point A and B
The function is

f(x) = x² -6x +a

(where "a" is a parameter)

The equation tangent line to the function at point A is
y = -2x +1

The equation tangent line to the function at point B
y = 2x -11

http://img864.imageshack.us/i/calc01.jpg/

The Attempt at a Solution

In order to find A I just need to set the function equal to the line equation at A, since this is the point where they meet. The problem is that "a" is a parameter. Until I know a I can't solve this equation. I can see "a" as an unknown, but then I got 1 equation and 2 unknowns.

I'm rather confused as to how to find this point without "a" being defined to me.

Any help?

 

[I like Serena]

What's the derivative of f(x)?

To find the x coordinate of A you need f'(x) = -2.
Can you solve that?

 

[Femme_physics]

How will the derivative of f(x) help me find that point? Do I set the derivative of the sad-smile parabola's function equal to the slope function?

To find the x coordinate of A you need f'(x) = -2.

Hmm, wait, the derivative of the slope function gives you the point A?

 

[Amok]

What's the derivative of f(x)?

To find the x coordinate of A you need f'(x) = -2.
Can you solve that?

Isn't this precalculus math? Have you taken calculus Femme Physique?

How will the derivative of f(x) help me find that point? Do I set the derivative of the sad-smile parabola's function equal to the slope function?
Hmm, wait, the derivative of the slope function gives you the point A?

The derivative of your function at a certain point gives you the slope of the tangent at that point. The slope of your first tangent line is -2, so if you set f'(x) = -2 and solve for x, you should find the abscissa of point A.

 

[I like Serena]

How will the derivative of f(x) help me find that point? Do I set the derivative of the sad-smile parabola's function equal to the slope function?



Hmm, wait, the derivative of the slope function gives you the point A?

The derivative of a function is the slope of that function.

In point A the slope of f(x) must be equal to the slope of the given tangent line in A.
This gives you 1 equation with 1 unknown...

 

[Femme_physics]

Isn't this precalculus math? Have you taken calculus Femme Physique?

"Femme_Physique"? *chuckles* :D

I know calc, I just thought it's a problem where I don't have to use calculus.

The derivative of your function at a certain point gives you the slope of the tangent at that point. The slope of your first tangent line is -2, so if you set f'(x) = -2 and solve for x, you should find the abscissa of point A.

Oh, I see, it makes sense actually, thanks! I'll try it right now :)

And thanks to you ILS as always!

 

[Mentallic]

You don't need calculus for this problem. Think about this, if you try to equate a line and a parabola (to see where they cut), you'll end up solving a quadratic equation. There could either be 2 solutions, 1 solution, or none at all. What do each of these mean in terms of the parabola and line?

 

[Femme_physics]

You don't need calculus for this problem. Think about this, if you try to equate a line and a parabola (to see where they cut), you'll end up solving a quadratic equation.

Yes-- but I don't have "a", "a" is a parameter.

I was able to solve it only using calculus.

 

[Mentallic]

Yes I know that, you need to find a first before you can find the coordinates of intersection A and B. Just try to equate one of the lines with the parabola, and take into consideration when finding the roots (which will be in terms of a) that the line is tangent to it.

 

[Femme_physics]

Makes sense. "a" is how much the function moves to the right, right? And I can also find a but finding the minima point, correct?

Hmm..wait, I turn up with a = 3, but the answer book says 5. I did do what you said

-2x+1 = 2x -11
4x = 12
x = 3

 

[Mentallic]

No, a is how much it moves up and down. If you have some parabola y_1=ax^2+bx+c then the parabola y_2=ax^2+bx+c+1=y_1+1 will move everything up 1 more unit.

You found where the two lines intersect each other by equating them together, instead equate the parabola and one of the lines.

 

[Femme_physics]

But if I equate the parabola's function and one of the slope's function, I still have 1 equation and 2 unknowns (x and a are unknown)

 

[Mentallic]

But if I equate the parabola's function and one of the slope's function, I still have 1 equation and 2 unknowns (x and a are unknown)

I've already explained this. You'll be solving for x (the answer will be in terms of a) and then you need to take one more detail into consideration:

You don't need calculus for this problem. Think about this, if you try to equate a line and a parabola (to see where they cut), you'll end up solving a quadratic equation. There could either be 2 solutions, 1 solution, or none at all. What do each of these mean in terms of the parabola and line?

 

[I like Serena]

The equation you would have is:

x² - 6x + a = -2x + 1

or:
x² - 4x + (a - 1) = 0

This is a quadratic equation in x (which includes an "a", but we'll ignore that for now).

The solution is (see also http://en.wikipedia.org/wiki/Quadratic_equation):

x=\frac {-(-4) \pm \sqrt {(-4)^2 - 4 \cdot 1 \cdot (a-1)} } {2 \cdot 1}

This yields 2 solutions, but only if the sqrt-part is greater than zero.
In particular if the sqrt-part equals zero, there is exactly 1 solution, which is what we're looking for.

So we have:
(-4)^2 - 4 \cdot 1 \cdot (a-1)} = 0

From this equation you can find a, and then you can also find x.

 

[Femme_physics]

I kinda get stuck here, unsure how to take a square root of a number with an unknown

 

[I like Serena]

I kinda get stuck here, unsure how to take a square root of a number with an unknown

That's right, you cannot take the square root of a number with an unknown.

The "trick" is that you know something else: there must be only 1 solution.

Try it for instance with a = 0 and also with a = 2.
Which (and in particular: how many) solutions do you get?

[edit]btw, I just saw that your calculation that came out to (12a - 12) is not quite right.
I guess we can get to that in a minute.[/edit]

 

[Femme_physics]

I plug (-4)²-4 * 1 and I get 12

So from there it's 12(a-1) = 12a -12

That's for the numerator under the square root

 

[I like Serena]

I plug (-4)²-4 * 1 and I get 12

So from there it's 12(a-1) = 12a -12

That's for the numerator under the square root

There's a rule that multiplying comes before adding/subtracting.

So (-4)² - 4 * 1 * (a - 1) = (-4)² - (4 * 1 * (a - 1))
(Notice that I've added a couple of parentheses to emphasize the order of evaluation.)

This comes out to:
16 - (4 * (a - 1))

Since you have an unknown, you can't calculate e.g. (a - 1) yet, but what you can do is rewrite the expression, eliminating the parentheses.

That is:
16 - (4 * a - 4 * 1) = 16 - (4a - 4)

[edit]That is because:
2 * (3 + 4) = (2 * 3) + (2 * 4)
Try it if you don't believe me! :)
[/edit]

 

[Mentallic]

Femme_physics you should revise your algebra!

 

[Femme_physics]

*smacks forehead* It's the calculator's fault! I mean, the way I tried to solve it using a calculator instead of writing everything down.

I'll try it again then. Thanks :)

 

[Femme_physics]

So, there it is. I'm still getting stuck with the same concept that I can't take the square root of an unknown.

That's right, you cannot take the square root of a number with an unknown.

The "trick" is that you know something else: there must be only 1 solution.

Try it for instance with a = 0 and also with a = 2.
Which (and in particular: how many) solutions do you get?

Hm. But I can't just set "a" to whatever I want, right?

 

[I like Serena]

So, there it is. I'm still getting stuck with the same concept that I can't take the square root of an unknown.

I'm going to take this one step at a time.

You write:
16 - 4(a - 1)

and then you make from this:
16 - 4a - 1

This is not right.
If you for instance fill in "a = 2", you'll get:
16 - 4(2 - 1) = 16 - 4(1) = 12
respectively:
16 - 4 x 2 - 1 = 16 - 8 - 1 = 7

You need to find something that if you fill in a value for "a" the results before and after match.

Hm. But I can't just set "a" to whatever I want, right?

Can you set "a" to whatever you want?
Well, yes and no.

Yes, you can set "a" to any value you want, and see what comes out.
But ultimately "a" must have a value such that the parabola will "touch" the lines in points A and B.

 

[Femme_physics]

Okay, so I was rush at getting to the next step and didn't solve correctly for
16 - 4(a - 1)

Clearly it's 16 -4a +4
Which then becomes
20 -4a

You need to find something that if you fill in a value for "a" the results before and after match.

So this is a guessing game? I just start from a=1 till a=1000 until I find something that matches?

 

[I like Serena]

Okay, so I was rush at getting to the next step and didn't solve correctly for
16 - 4(a - 1)

Clearly it's 16 -4a +4
Which then becomes
20 -4a

So this is a guessing game? I just start from a=1 till a=1000 until I find something that matches?

All right!
Let's make it a guessing game! :)
Take a guess for "a" between 1 and 1000.
Can you give me a number?

[edit]I pick a = 1. What's yours?[/edit]

[edit2]Can you give me the solutions for x for your choice?[/edit2]

 

[Femme_physics]

Okay, I gave it a shot, and after three random guesses it appears that a=5, where the whole shebang under the square root goes to zero, that gives the same 2 results.

So I'm always supposed to get the whole shebang under the square root to get to 0 to find a?

 

[I like Serena]

Okay, I gave it a shot, and after three random guesses it appears that a=5, where the whole shebang under the square root goes to zero, that gives the same 2 results.

So I'm always supposed to get the whole shebang under the square root to get to 0 to find a?

Only if you want the equation to have exactly 1 solution, which now we do.

It's a kind of special case in which you can solve 2 unknowns from 1 equation, but only with the extra information that the parabola "touches" the line.

 

[Femme_physics]

It's a kind of special case in which you can solve 2 unknowns from 1 equation, but only with the extra information that the parabola "touches" the line.

Woahhhhh...I'm on math drug...

Many, many thanks for getting me there ILS!

 

[I like Serena]

Earlier, you wrote this.

Yes-- but I don't have "a", "a" is a parameter.

I was able to solve it only using calculus.

Did you get the same values for x and a (using calculus)?

 

[Femme_physics]

Well, I just needed to find a, not x! I'm not sure what you mean

By the way, now it's really getting into calculus, for the last question I need to find the area that's caged above the slopes but below the function (graphed in the original upload).

I persume I take the integral of the parabola's function from 2 to 4, then reduce it by the integrals of the two slopes from 2 to 4. Does it make sense?

 

[I like Serena]

Well, I just needed to find a, not x! I'm not sure what you mean

The problem statement asks for the coordinates of A and B.
You have just not only found a = 5, but also x = 2 for point A.
You're not entirely finished yet, because you also need the y-coordinate of A, and the coordinates of B.


However, my original response at the start of the thread was:

To find the x coordinate of A you need f'(x) = -2.

In point A the slope of f(x) must be equal to the slope of the given tangent line in A.
This gives you 1 equation with 1 unknown...

This is first step in the "calculus" way of solving the problem using the derivative, which is actually easier.
It's before you were set on the course to intersect a parabola with a line, triggered by the comment of Mentalic.

Since you're learning calculus this seems to be a good place to practice it too. :)

By the way, now it's really getting into calculus, for the last question I need to find the area that's caged above the slopes but below the function (graphed in the original upload).

I persume I take the integral of the parabola's function from 2 to 4, then reduce it by the integrals of the two slopes from 2 to 4. Does it make sense?

Yes, that makes perfect sense!

Take note however that you need to take the integral of the first slope from 2 to 3, and the integral of the second slope from 3 to 4.

 

[Femme_physics]

The problem statement asks for the coordinates of A and B.
You have just not only found a = 5, but also x = 2 for point A.
You're not entirely finished yet, because you also need the y-coordinate of A, and the coordinates of B.

Oh yea, that's easy. I actually got
A (x = 2)

So in that case, I plug 2 and get

Y = -3

For B I plug in 4, and I get
Y = -3

So

A (2,-3)
B (4, -3)

This is first step in the "calculus" way of solving the problem using the derivative, which is actually easier.
It's before you were set on the course to intersect a parabola with a line, triggered by the comment of Mentalic.

Since you're learning calculus this seems to be a good place to practice it too. :)

Oh yes, I didn't actually do the intersection, as you can see from my attached full scan (answer to question 2)

Take note however that you need to take the integral of the first slope from 2 to 3, and the integral of the second slope from 3 to 4.

Eep! Sound complex, as I only solved for 2 integrals at best so far, now I have 3! But, going to give it a shot, for sure! :)

Many thanks ILS!

 

[I like Serena]

Oh yea, that's easy. I actually got
A (x = 2)

So in that case, I plug 2 and get

Y = -3

For B I plug in 4, and I get
Y = -3

So

A (2,-3)
B (4, -3)

Oh yes, I didn't actually do the intersection, as you can see from my attached full scan (answer to question 2)

Uhhh, yes, that's entirely correct.

And sorry to keep bugging you, but could you also solve "a" this way?
That is, since your scan suddenly switches from the calculus-derivative-method to the algebra-parabola-intersection-method.
Could you have calculated "a" in a different manner using your previous results?

Eep! Sound complex, as I only solved for 2 integrals at best so far, now I have 3! But, going to give it a shot, for sure! :)

Many thanks ILS!

Always!

 

[Femme_physics]

And sorry to keep bugging you, but could you also solve "a" this way?

You're practicing my math, it's me who's bugging you if anything!

That is, since your scan suddenly switches from the calculus-derivative-method to the algebra-parabola-intersection-method. Could you have calculated "a" in a different manner using your previous results?

Oh. I see. I hardly knew how to find it through non-calculus! (setting the whole thing under the square root equal to zero!) Thanks to you I now do. Is there another method? How?

 

[I like Serena]

Oh. I see. I hardly knew how to find it through non-calculus! (setting the whole thing under the square root equal to zero!) Thanks to you I now do. Is there another method? How?

The equation of your parabola is:
f(x) = x² -6x +a

You found that A is (2, -3) using calculus.
This is a point on the parabola.

Can you combine that?
And solve for a?

 

[HallsofIvy]

In order to intersect you must have x^2- 6x+ a= -2x+ 1 so that x must satisfy x^2- 4x+ a- 1= 0.

In order that the line be tangent to the parabola, that value of x must be a double root. That is, we must have x^2- 4x+ a- 1= (x- x_0)^2.

Complete the square in x^2- 4x+ a- 1. a must be such this is as perfect square.

 

[Femme_physics]

The equation of your parabola is:
f(x) = x² -6x +a

You found that A is (2, -3) using calculus.
This is a point on the parabola.

Can you combine that?
And solve for a?

Let me see,

So
Ay = -3
Ax = 2

(this is starting to sound like mechanics, but bear with me lol)

So I plug that into the parabola function and I get

-3 = 4 -12 +a

a = 5

YESSSSSSSSSSSS!

Is that it? :D Has to be? Brilliant!

Tons easier, too! I wished we used that method to begin with!

In order to intersect you must have x^2- 6x+ a= -2x+ 1 so that x must satisfy x^2- 4x+ a- 1= 0.

In order that the line be tangent to the parabola, that value of x must be a double root. That is, we must have x^2- 4x+ a- 1= (x- x_0)^2.

Complete the square in x^2- 4x+ a- 1. a must be such this is as perfect square.

Oh, I believe I already found it using this route (posted above), but thanks :)

 

[I like Serena]

YESSSSSSSSSSSS!

Is that it? :D Has to be? Brilliant!

Tons easier, too! I wished we used that method to begin with!)

That's it!

(And you're making me feel good about myself! )

 

[Femme_physics]

You're the one who's empowering my math, you're making me feel AWESOME about myself! Thank you :D

 

[Femme_physics]

Hm, someone I'm not getting the area result with the notion of integrating the 3 functions...can you tell me what I'm doing wrong?

Answer book says 2/3

 

[I like Serena]

Hm, someone I'm not getting the area result with the notion of integrating the 3 functions...can you tell me what I'm doing wrong?

Answer book says 2/3

Before I answer your question, I'd like you to consider what you are doing.

I think you already know that the integral of a function measures the surface between the function and the x-axis.
With a function below the x-axis, the integral will come out to a negative value.

Can you estimate what those surfaces are for the 3 integrals you have?
Btw, a typical estimation would be based on a rectangle that more or less matches the surface.

 

[Femme_physics]

With a function below the x-axis, the integral will come out to a negative value.

Yes, I see I forgot to add minus to the eight in the last calculation, but even then it's wrong

Can you estimate what those surfaces are for the 3 integrals you have?

I guess I can make a triangle if I find the critical point (I already know it's 3, actually) and that would give an "estimate" without the meniscus. I'm not sure how it helps me get the real value, though

 

[I like Serena]

Yes, I see I forgot to add minus to the eight in the last calculation, but even then it's wrong

I guess I can make a triangle if I find the critical point (I already know it's 3, actually) and that would give an "estimate" without the meniscus. I'm not sure how it helps me get the real value, though

Let's take the third integral as an example.
The corresponding rectangle would have width 1 and a height between 3 and 4.
So the integral should come out between -3 and -4.
Instead, what you have is +2.

Did you make a mistake copying your formula or something? ;)

 

[Femme_physics]

http://img705.imageshack.us/img705/2505/mistakewhere.jpg

BTW can a mod move this to the calculus section?

 

[I like Serena]

In your scan you have calculated the third integral as: x² - 11x which is correct.
Which formula did you use when you substituted 3 and 4?

Btw, in your first integral you did use the proper formula, but apparently you made a mistake with your calculator?
Note that the estimation of the first integral would be a rectangle with width 2 and height between 3 and 4. So the first integral should come out between -6 and -8, but *not* exactly -8.

 

[Femme_physics]

Btw, in your first integral you did use the proper formula, but apparently you made a mistake with your calculator?

Yes! Sharp of you. It's -7.3333.

But, it's still off the mark

Which formula did you use when you substituted 3 and 4?

The ascending slope
i.e.
2x-11
whose integral is x² -11x

 

[I like Serena]

The ascending slope
i.e.
2x-11
whose integral is x² -11x

In your calculation of the third integral you used 2x-11 when you should have used x² -11x.

Usage of y=2x-11 would give you the difference in the y values at x = 3 and x = 4, which is indeed +2, but then you need the surface... :)

 

[Femme_physics]

Got it!

I now see all the stupid mistakes I've made! I actually didn't take the integral but the function when I did the area calcultion of the 2 slopes! *slaps forehead*.

As you can see, I haven't taken areas two many times. I'm self-studying entirely for an external math exam that's mandatory for me if I want to keep studying engineering.

You're unbelievable for helping me through all of that ILS! Thank you so much :D

 

[I like Serena]

Got it!

I now see all the stupid mistakes I've made! I actually didn't take the integral but the function when I did the area calcultion of the 2 slopes! *slaps forehead*.

As you can see, I haven't taken areas two many times. I'm self-studying entirely for an external math exam that's mandatory for me if I want to keep studying engineering.

You're unbelievable for helping me through all of that ILS! Thank you so much :D

Wait!
Does that mean this thread has come to an end?
That we must move on to a new and uncertain thread?
I don't want it to end!

 

[Femme_physics]

LMAO! :D

The exercise has indeed sadly come to an end, but my help-seeking ways and scanning days are just beginning ;)

I have a bunch more! I just make it a rule not to ask 3 questions simultaneously.

I'll be hoping to see you stopping by! :D Always happy to see you've replied. Thanks. Learning so much from you.