A particle is moving along the parabola 4y = (x + 2)^2 in such a way that
its x-coordinate is increasing at a constant rate of 2 units per second. How
fast is the particle's distance to the point (-2, 0) changing at the moment
that the particle is at the point (2, 4)?
Let the point be at the position P(x,y)
The parabola has vertex at (-2,0), so the question is how fast is P moving away from the vertex.
Let the distance be D
D^2 = (x+2)^2 + y^2
= (x+2)^2 + (x+2)^4/16
2D dD/dt = 2(x+2) dx/dt + (1/4)(x+2)^3 dx/dt
dD/dt = [2(x+2) dx/dt + (1/4)(x+2)^3 dx/dt]/(2D)
when x = 2
D^2 = 16 + 256/16
= 32
D = √32
dD/dt = [2(4)(2) + (1/4(64)(2)]/(2√32)
= 24/√32 = 6/√2 = appr. 4.243 units per second
check my arithmetic
To find how fast the particle's distance to the point (-2, 0) is changing, we can use the concept of related rates.
Let's denote the distance between the particle and the point (-2, 0) as D. We want to find dD/dt, the rate at which D is changing with respect to time.
First, let's find an equation that relates D to the coordinates of the particle (x, y). The distance D is the hypotenuse of a right triangle formed by the line segment connecting the particle and the point (-2, 0), and the vertical line segment connecting the particle and the x-axis.
Using the Pythagorean theorem, we can write:
D^2 = (x - (-2))^2 + (y - 0)^2
D^2 = (x + 2)^2 + y^2
Next, we need to express the variables x and y in terms of time t, since they are changing with respect to time. We are given that the x-coordinate is increasing at a constant rate of 2 units per second. So, we can write:
dx/dt = 2
Now, let's differentiate both sides of the equation D^2 = (x + 2)^2 + y^2 with respect to time t using implicit differentiation:
d(D^2)/dt = d((x + 2)^2 + y^2)/dt
Using the chain rule, we have:
2D * dD/dt = 2(x + 2) * dx/dt + 2y * dy/dt
Since we are interested in finding dD/dt, let's rearrange the equation:
dD/dt = ((x + 2) * dx/dt + y * dy/dt) / D
We're given that the particle is at the point (2, 4), so we can substitute x = 2 and y = 4 into the equation:
dD/dt = ((2 + 2) * 2 + 4 * dy/dt) / D
Now, we need to find dy/dt, the rate at which y is changing with respect to time. To do this, we can differentiate the equation of the parabola 4y = (x + 2)^2 implicitly:
d(4y)/dt = d((x + 2)^2)/dt
4 * dy/dt = 2(x + 2) * dx/dt
Substituting dx/dt = 2, we have:
4 * dy/dt = 2(2 + 2)
Simplifying, we get:
4 * dy/dt = 8
Dividing both sides by 4, we find:
dy/dt = 2
Now, we can substitute dx/dt = 2, x = 2, y = 4, and dy/dt = 2 into the equation for dD/dt:
dD/dt = ((2 + 2) * 2 + 4 * 2) / D
dD/dt = (4 * 2 + 4 * 2) / D
Simplifying, we find:
dD/dt = (8 + 8) / D
dD/dt = 16 / D
To find the value of D, we substitute x = 2 and y = 4 into the equation D^2 = (x + 2)^2 + y^2:
D^2 = (2 + 2)^2 + 4^2
D^2 = 16 + 16
D^2 = 32
Taking the square root of both sides, we find:
D = √32
D = 4√2
Finally, we can substitute D = 4√2 into the equation for dD/dt:
dD/dt = 16 / (4√2)
Simplifying, we get:
dD/dt = 4 / √2
dD/dt = 4√2 / 2
dD/dt = 2√2
Therefore, the particle's distance to the point (-2, 0) is changing at a rate of 2√2 units per second at the moment when the particle is at the point (2, 4).