A particle is moving along the curve y=4((3x+1)^.5). As the particle passes through the point (5,16) 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.
To find the rate of change of the distance between the particle and the origin, we need to use the concept of the Pythagorean theorem. The distance between two points (x₁, y₁) and (x₂, y₂) can be calculated using the formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
In this case, we want to find the rate of change of distance with respect to time:
(d/dt) = (d/dx) * (dx/dt) + (d/dy) * (dy/dt)
First, let's find the values of x and y when the particle passes through the point (5,16). Plugging these values into the equation for y:
16 = 4((3*5 + 1)^(1/2))
Simplifying:
16 = 4(16^(1/2))
16 = 4(4)
16 = 16
So, when the particle passes through the point (5, 16), the x-coordinate remains constant at 5.
Next, we are given that the x-coordinate is increasing at a rate of 2 units per second: (dx/dt) = 2.
Taking the derivative of the equation of the curve y = 4((3x + 1)^.5) with respect to x:
(dy/dx) = 4 * ((3x + 1)^.5)'}
Using the chain rule:
(dy/dx) = 4 * (1/2) * (3x + 1)^(-1/2) * 3
Simplifying:
(dy/dx) = 6(3x + 1)^(-1/2)
Now, let's plug in the value of x as 5:
(dy/dx) = 6(3(5) + 1)^(-1/2)
= 6(16)^(-1/2)
= 6/4
= 3/2
Finally, substitute the values we found into the rate of change formula:
(d/dt) = (d/dx) * (dx/dt) + (d/dy) * (dy/dt)
= 2 * (3/2)
= 3
Therefore, the rate of change of the distance from the particle to the origin at the instant it passes through the point (5, 16) is 3 units per second.