Question

A particle is moving along the curve y= 4sqrt(5x+11) . As the particle passes through the point (5,24) , its -coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant

Answer 1

The rate of change of the distance from the particle to the origin at this instant is 3 units per second.

What is the rate of change?

The instantaneous rate of change is the rate of change at a particular instant.

A particle is moving along the curve

$y= 4\sqrt{5x+11} .$

The rate of change of y is given as:

dy / dx = 2

by differentiating both sides,

$\dfrac{dy}{dx} = 4\dfrac{1}{2\sqrt{5x+11} } 5\dfrac{dx}{dt} \\\\\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}$

From the question, we have:

(x, y) =  (5,24)

Substitute 5 for x and dy / dx = 2

$\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\\\5= \dfrac{10}{\sqrt{5(5)+11} }\dfrac{dx}{dt}\\\\\\\sqrt{5(5)+11} = 2\dfrac{dx}{dt}\\\\\\\dfrac{dx}{dt} = 6/ 2 = 3$

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:

brainly.com/question/13103052

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