Kelly Novak
AP Review 8 Problem #6
The function f is differentiable for all real numbers. The point (3, ¼) is on the graph of y=f(x), and the
slope at each point (x, y) on the graph is given by dy/dx = y2(6-2x).
(a.) Find d2y/dx2 and evaluate it at the point (3, ¼).
(b.) Find y= f(x) by solving the differential equation dy/dx= y2(6-2x) and f(3)= ¼
(a.):
We know dy/dx, so the first thing to do is get its derivative (d2y/dx2).
dy/dx=y2(6-2x)
Use the multiplication method to find the derivative and you get:
d2y/dx2 = 2yy’(6-2x)-2 y2
Now, there is a y’ in that equation, so in order to solve at (3, ¼), you have to first find dy/dx at
that point.
dy/dx= (¼)2[6-2(3)]
dy/dx= (1/16)(6-6)
dy/dx= (1/16)(0)
dy/dx= 0
So now that we have y’ at that point, we can plug all of our values back into the second
derivative equation.
d2y/dx2= 2(¼) (0)[6-2(3)]-2(¼)2
d2y/dx2= 0-2(1/16)
d2y/dx2= -1/8
The second derivative at (3, ¼) is -1/8.
(b.):
dy/dx=y2(6-2x)
dy/y2=6-2x dx
Kelly Novak
∫ 𝑦 −2 𝑑𝑦=∫ 6 − 2𝑥 𝑑𝑥
−𝑦 −1=6x-x2 + C
y-1=x2-6x-C
1/y= x2-6x-C
Y= 1/(x2-6x-C)…=f(x)
Now plug in the x- & y-values to find C.
¼ =1/[(3)2-6(3)-C]
¼ =1/(9-18-C)
4=9-18-C
4= -9-C
13= -C
-13=C
Now plug in the C value into the solved differentially equation! :D
Y= 1/(x2-6x+13)