Suppose y= √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= 5, find dx/dt when x=12. 

Solution:

y = √2x + 1 [Given]

Differentiating both sides with respect to t

dy/dt = dx/dt/ √(2x + 1)

(a) if dx/dt= 3, we have to find dy/dt when x = 4

dy/dt = 3/√(2(4) + 1)

dy/dt = 3/√(9)

dy/dt = 3/√9

dy/dt = 3/3

dy/dt = 1

(b) if dy/dt= 5, we have to find dx/dt when x = 12. 

5 = dx/dt/ √(2(12) + 1)

5 × √(24 + 1) = dx/dt

dx/dt = 5 × √25

dx/dt = 5 × 5

dx/dt = 25

Therefore, the values of dy/dt is 1 and dx/dt is 25.

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Suppose y= where x and y are functions of t. (a) if dx/dt= 3, find dy/dt when x=4. (b) if dy/dt= 5, find dx/dt when x=12. 

Summary:

Suppose √2x + 1 where x and y are functions of t.

(a) if dx/dt= 3, dy/dt is 1 when x = 4. (b) if dy/dt= 5, dx/dt is 25 when x = 12.