A particle is moving along the curve y = 2 √{3 x + 7}. As the particle passes through the point (3, 8), its x-coordinate increases at a rate of 5 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 from the particle to the origin at the instant when it passes through the point (3, 8), we can use the concept of the Pythagorean theorem.
The distance from a point (x, y) to the origin (0, 0) is given by the formula:
distance = √(x² + y²).
We need to find the rate of change of this distance. Let's call this rate dy/dx (the derivative of y with respect to x).
First, let's find the derivative of y = 2 √(3x + 7) with respect to x:
dy/dx = d/dx (2 √(3x + 7))
To find this derivative, we can use the chain rule. Let's substitute u = 3x + 7 and differentiate:
dy/dx = d/dx (2 √u)
= 2 * (1/2) * u^(-1/2) * du/dx
= (1/√u) * du/dx
= (1/√(3x + 7)) * d/dx (3x + 7)
= (1/√(3x + 7)) * 3
= 3/√(3x + 7)
Now we need to evaluate this derivative at the point (3, 8):
dy/dx = 3/√(3(3) + 7)
= 3/√(9 + 7)
= 3/√16
= 3/4
So, the rate of change of y with respect to x when the particle passes through the point (3, 8) is 3/4.
Now, we can find the rate of change of the distance from the particle to the origin using the Pythagorean theorem:
distance = √(x² + y²)
The rate of change of the distance with respect to time (dt) is given by the chain rule:
d(distance)/dt = d(distance)/dx * dx/dt
Given that dx/dt = 5 units per second, we can substitute this value into the equation:
d(distance)/dt = (dy/dx) * (dx/dt)
= (3/4) * 5
= 15/4
Therefore, the rate of change of the distance from the particle to the origin at the instant when it passes through the point (3, 8) is 15/4 units per second.
To find the rate of change of the distance from the particle to the origin, we can use the distance formula:
Distance = √(x^2 + y^2)
Differentiating both sides with respect to time, we get:
d(Distance)/dt = d(√(x^2 + y^2))/dt
Since we know that dx/dt = 5 (the x-coordinate is increasing at a rate of 5 units per second), we need to find dy/dt (the rate of change of y with respect to time) to evaluate the derivative above.
To find dy/dt, we can rearrange the equation y = 2√(3x + 7) as:
y^2 = 12x + 28
Differentiating both sides with respect to time, we get:
2yy' = 12x'
Now, we can substitute the given information: at the point (3,8), x' = 5 (dx/dt = 5) and y = 8. Plugging these values into the equation above, we can solve for y':
2(8)y' = 12(5)
16y' = 60
y' = 60/16
y' = 15/4
Now, we can substitute the values of x' and y' into our original expression for the rate of change of the distance:
d(Distance)/dt = d(√(x^2 + y^2))/dt
d(Distance)/dt = d(√(x^2 + (2√(3x + 7))^2))/dt
d(Distance)/dt = d(√(x^2 + 12x + 28))/dt
d(Distance)/dt = (1/2)(x^2 + 12x + 28)^(-1/2)(2x + 12)(dx/dt)
Plugging in the values of x and dx/dt at the point (3,8), we get:
d(Distance)/dt = (1/2)(3^2 + 12(3) + 28)^(-1/2)(2(3) + 12)(5)
d(Distance)/dt = (1/2)(9 + 36 + 28)^(-1/2)(18 + 12)(5)
d(Distance)/dt = (1/2)(73)^(-1/2)(30)(5)
d(Distance)/dt = (15/2)(1/√73)
Therefore, the rate of change of the distance from the particle to the origin at the instant when it passes through the point (3,8) is (15/2)(1/√73) units per second.