Question

Suppose y = sqrt 2x+1, where x and y are functions of t.

(a) If dx/dt = 9, find dy/dt when x = 4.
(b) If dy/dt = 3, find dx/dt when x = 40.

Answer 1

If y = √(2x + 1), then differentiating both sides implicitly with respect to t gives

dy/dt = 1/2 • 1/√(2x + 1) • 2 • dx/dt = 1/√(2x + 1) • dx/dt

(a) If dx/dt = 9 and x = 4, then

dy/dt = 1/√(2•4 + 1) • 9

dy/dt = 1/√(8 + 1) • 9

dy/dt = 1/√9 • 9

dy/dt = 9/3

dy/dt = 3

(b) If dy/dt = 3 and x = 40, then

3 = 1/√(2•40 + 1) • dx/dt

3 = 1/√(80 + 1) • dx/dt

3 = 1/√81 • dx/dt

3 = 1/9 • dx/dt

dx/dt = 27