AP Calculus BC 5.12: Implicitly Defined Functions
Practise AP Calculus BC 5.12 by differentiating relations involving x and y, finding second derivatives, and using their signs to justify concavity.
- Syllabus
- Effective Fall 2025
- Course
- AP Calculus BC
Practise AP Calculus BC 5.12 by differentiating relations involving x and y, finding second derivatives, and using their signs to justify concavity.
Given the following differential equation .
(b) Find the second derivative, dx2d2y, in terms of x and y. The region in
the x y-plane where all the solution curves to the differential
equation are concave down can be expressed as a linear inequality.
Find this region.
(b)
art fraction numerator d squared end exponent y over
denominator d x squared end exponent end fraction equals 1
minus 2 start fraction numerator d y over denominator d x end
fraction equals 1 minus 2 left parenthesis x minus 2 y plus 1
right parenthesis equals 4 y minus 2 x minus 1
The solution curves will be concave down when
- start fraction numerator d squared y over denominator d x
squared end fraction less than 0
4 y minus 2 x minus 1 less than 0 rightwards double arrow y less than 1 over 2
x plus one-fourth