SOLUTION: Suppose y = sqrt 2x + 1 where x and y are functions of t.
a) If dx/dt = 3,find dy/dt when x = 4.
(b) If dy/dt = 4,find dx/dt when x = 12.
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-> SOLUTION: Suppose y = sqrt 2x + 1 where x and y are functions of t.
a) If dx/dt = 3,find dy/dt when x = 4.
(b) If dy/dt = 4,find dx/dt when x = 12.
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You can put this solution on YOUR website! y = sqrt 2x + 1 where x and y are functions of t
square both sides of =
y^2 = 2x + 1
Now differentiate implicitly with respect to t:
2y*dy/dt = 2dx/dt
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a) If dx/dt = 3,find dy/dt when x = 4
from original equation
y = sqrt(2*4 + 1) = 3
now use our differential equation
(2*3)*dy/dt = 2*3
dy/dt = 6/6 = 1
***************************************************
(b) If dy/dt = 4,find dx/dt when x = 12
from original equation
y = sqrt(2*12 + 1) = 5
now use our differential equation
(2*5)*4 = 2*dx/dt
40 = 2*dx/dt
dx/dt = 20
