Question

Ap Calculus BC 2002 Form B Question 5?

Answer 1

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!