A particle is moving along the curve y=5 sqrt (2x+6). As the particle passes through the point (5,20 , 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, we need to find the distance formula between two points in the Cartesian coordinate system.
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
In this case, one of the points is the origin (0, 0), and the other point is the position of the particle at the given instant (5, 20). Let's call the position of the particle (x, y).
So, the distance from the particle to the origin is:
d = √[(x - 0)^2 + (y - 0)^2]
= √[x^2 + y^2]
To find the rate of change of the distance, we need to differentiate the distance formula with respect to time, since the x-coordinate of the particle is changing with time.
Differentiating both sides of the equation with respect to time (t), we get:
d/dt = d/dt √[x^2 + y^2]
To find d/dt √[x^2 + y^2], we need to apply the chain rule. Let's calculate the derivatives step by step:
1. Differentiating √[x^2 + y^2] with respect to t:
d/dt √[x^2 + y^2] = (1/2) * (x^2 + y^2)^(-1/2) * d/dt (x^2 + y^2)
2. Differentiating (x^2 + y^2) with respect to t:
d/dt (x^2 + y^2) = d/dt (x^2) + d/dt (y^2)
Since we are given that the x-coordinate increases at a rate of 5 units per second, we can substitute dx/dt = 5.
So, d/dt (x^2) = 2x * dx/dt = 2x * 5 = 10x
3. Differentiating (y^2) with respect to t:
d/dt (y^2) = 2y * dy/dt
To find dy/dt, we can differentiate the given equation y = 5 sqrt(2x + 6) with respect to t:
dy/dt = d/dt (5 sqrt(2x + 6))
Again, we apply the chain rule to differentiate sqrt(2x + 6) with respect to t:
dy/dt = (5/2) * (2x + 6)^(-1/2) * d/dt (2x + 6)
Since we are given that the x-coordinate increases at a rate of 5 units per second, we can substitute dx/dt = 5.
So, d/dt (2x + 6) = 2 * dx/dt = 2 * 5 = 10
Substituting these results back into the differentiation of (y^2) with respect to t:
d/dt (y^2) = 2y * dy/dt = 2y * [(5/2) * (2x + 6)^(-1/2) * 10] = 2y * (5/2) * (2x + 6)^(-1/2) * 10
Now we have all the derivatives, so we can calculate d/dt √[x^2 + y^2]:
d/dt √[x^2 + y^2] = (1/2) * (x^2 + y^2)^(-1/2) * (10x + 2y * (5/2) * (2x + 6)^(-1/2) * 10)
Simplifying:
d/dt √[x^2 + y^2] = 5(x^2 + y * (2x + 6)^(-1/2)) / √[x^2 + y^2]
Finally, we substitute the given values at the point (5, 20):
d/dt √[5^2 + (20 * (2 * 5 + 6)^(-1/2))] / √[5^2 + 20^2]
Calculating the expression will give us the rate of change of the distance from the particle to the origin at this instant.
y = 5√(2x+6)
dy/dt = 5/√(2x+6) dx/dt
d^2 = x^2+y^2
d dd/dt = x dx/dt + y dy/dt
at (5,20), d=5√17, dy/dt = 5/4
5√17 dd/dt = 5*5 + 20*5/4
dd/dt = (25+25)/(5√17) = 10/√17