The rate of change of the distance from the particle to the origin at this instant is 3 units per second.
The instantaneous rate of change is the rate of change at a particular instant.
A particle is moving along the curve
[tex]y= 4\sqrt{5x+11} .[/tex]
The rate of change of y is given as:
dy / dx = 2
by differentiating both sides,
[tex]\dfrac{dy}{dx} = 4\dfrac{1}{2\sqrt{5x+11} } 5\dfrac{dx}{dt} \\\\\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\[/tex]
From the question, we have:
(x, y) = (5,24)
Substitute 5 for x and dy / dx = 2
[tex]\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\\\5= \dfrac{10}{\sqrt{5(5)+11} }\dfrac{dx}{dt}\\\\\\\sqrt{5(5)+11} = 2\dfrac{dx}{dt}\\\\\\\dfrac{dx}{dt} = 6/ 2 = 3[/tex]
Hence, the rate of change of the distance from the particle to the origin at this instant is 3 units per second.
Read more about rates of change at:
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