Ap Calculus BC 2002 Form B Question 5?

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1 Answer
Apr 28, 2018

a) If #y = -2# is tangent to the graph, it means that the slope of that particular tangent is #0#. Thus:

#0 = (3- x)/y#

We also know that #y = -2#. Now all we must do is solve for #x# which is pretty simple.

#0 = (3- x)/(-2)#

#x = 3#

To determine the nature of this critical point, we must determine the second derivative.

#(d^2y)/(dx^2) = (-1(y) - (3- x)(dy/dx))/y^2#

#(d^2y)/(dx^2) = (-y - (3- x)(3 - x)/y)/y^2#

#(d^2y)/(dx^2) = ( -y - (3 - x)^2/y)/(y^2)#

#(d^2y)/(dx^2) = (-y^2 - (3 -x)^2)/y^3#

Since this #(d^2y)/(dx^2)# is positive at the point #(3, -2)#, this is a minimum (remember a minimum is always concave up and a maximum always concave down).

b) This is a classic differential equation solving problem. Start by separating the x and the y.

#dy/dx(y)= 3 - x#

#dy(y) = 3 - x dx#

Integrate both sides.

#int y dy = int 3 - x dx#

#1/2y^2 = -1/2x^2 + 3x + C#

Never forget the constant of integration has to be included before you solve for #y#. If you forget #C# when solving this type of problem on the exam, they will only be able to give you a maximum of #3/6# marks! It's ok to rewrite as a different letter before you take the square root.

#1/2y^2 = -1/2x^2 + 3x+ C#

Now solve for #C#.

#1/2(-4)^2 = -1/2(6^2) + 3(6) + C#

#8 + 18 - 18 = C#

#C = 8#

It now follows that

#1/2y^2 = -1/2x^2 + 3x + 8#

#y^2 = -x^2 + 6x + 16#

#y = +- sqrt(-x^2 + 6x +16)#

But since the initial condition has #y# negative, we only keep the negative sign.

#y = - sqrt(6x - x^2 + 16)#

Hopefully this helps!