Expert: Paul Klarreich - 2/24/2006

Question
Hi,
I hope I'm finding you in the best of health.
http://apcentral.collegeboard.com/repository/b_sg_calculus_ab_02_11407.pdf
This above link will take you to the AP site by Collegeboard.
By question 5b, I don't understand why when they solved for y, y^2=6x-x^2+16,they only took the negative square root of 6x-x^2+16.Why did they exclude y=+sqrt(6x-x^2+16)?
And in question 5A(the question above 5b;sorry I went out of order), how did they know that x=3 is the x-coordinate of the point of tangency(y=-2)?Why did they set dy/dx=0.Where in the problem does it imply to set dy/dx equal to zero?
Thank you very much for your time.

Answer
Hi, Jeff,
You wrote:
Subject:  differential equations
Question:  Hi,
I hope I'm finding you in the best of health.
http://apcentral.collegeboard.com/repository/b_sg_calculus_ab_02_11407.pdf
This above link will take you to the AP site by Collegeboard.
------------------------------------
>> OK, I found it.

By question 5b, I don't understand why when they solved for y, y^2=6x-x^2+16,they only took the negative square root of 6x-x^2+16.Why did they exclude y=+sqrt(6x-x^2+16)?

>> Because you are specifying that you want the particular solution that gives y = - 4.  If you take the positive square root, you can't get that.

When you separate the variables, you have:

y dy = (3- x)dx, which integrates to:

y^2        x^2
--- = 3x - --- + c,  or
2          2

y^2 = 6x - x^2 + C

Now you want  g(6) = -4, and so you want an equation satisfied by x=6 and y=-4.  Obviously  y = +sqrt(...) won't do that.
-------------------
And in question 5A(the question above 5b;sorry I went out of order), how did they know that x=3 is the x-coordinate of the point of tangency(y=-2)?

Why did they set dy/dx=0.Where in the problem does it imply to set dy/dx equal to zero?

>> The file says:   
      


      

  
   
   
      

Oops -- that didn't come out good.  I will have to type it myself:

It says "the line  y = -2 is tangent to the graph"
That means the HORIZONTAL line y=-2.
That means the slope is zero at y=-2.
That means dy/dx = 0.
But dy/dx = (3-x)/y, which is a fraction.
A fraction is zero when the numerator is zero.
So 3-x = 0,  or  x = 3.