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

[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.