A particle is moving around the ellipse 4x^2+16y^2=64 . At any time t, its x- and y coordinates are given by x(t) = 4cos t and y(t)= 2sint . At what rate is the particle’s distance to the point (1,1) changing at any time t? At what rate is the distance changing when t=pi/6?

The distance z is

z^2 = (x-1)^2 + (y-1)^2

2z dz/dt = 2(x-1) dx/dt + 2(y-1) dy/dt
At t=π/6, z=√((4(√3/2)-1)^2+(2(1/2)-1)^2) = 2√3-1
Note that at t=π/6, the point on the curve is (2√3,1)

So, now we just have
dx/dt = -4sint = -2
dy/dt = 2cost = √3

(2√3-1) dz/dt = (2√3-1)(-2)+(1-1)(√3)
dz/dt = -2

Well, well, look at this particle going for a little joyride on that ellipse. Let's calculate its distance to the point (1,1) and see how fast it's changing.

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the lovely formula: distance = √((x₂ - x₁)² + (y₂ - y₁)²).

So in this case, the distance between the particle's coordinates (4cos(t), 2sin(t)) and (1,1) will be √((4cos(t) - 1)² + (2sin(t) - 1)²).

Now, let's differentiate this distance with respect to time t to find the rate at which it's changing. We'll use the chain rule for this.

(d/dt) [√((4cos(t) - 1)² + (2sin(t) - 1)²)] = (1/2) * [(4cos(t) - 1)² + (2sin(t) - 1)²]^(-1/2) * (d/dt) [(4cos(t) - 1)² + (2sin(t) - 1)²].

Oh dear, things are getting a bit messy. But don't worry, we're clowns, we can handle this. Let's simplify.

= (1/2) * [(4cos(t) - 1)² + (2sin(t) - 1)²]^(-1/2) * [2(4cos(t) - 1)(-sin(t)) + 2(2sin(t) - 1)(cos(t))].

Wow, that's a mouthful! But at least it's simple to evaluate. Now we need to find the rate at which the distance is changing when t = π/6. Let's plug in that special value and see what happens.

(d/dt) [√((4cos(t) - 1)² + (2sin(t) - 1)²)] = (1/2) * [(4cos(π/6) - 1)² + (2sin(π/6) - 1)²]^(-1/2) * [2(4cos(π/6) - 1)(-sin(π/6)) + 2(2sin(π/6) - 1)(cos(π/6))].

Calculating all those trigonometric functions, you'll find the exact rate at that specific time. It might look intimidating, but don’t worry, my clown friend, you can handle it!

To find the rate at which the particle's distance is changing with respect to time (t) at any time t, we can use the concept of the derivative. Specifically, we will use the chain rule to find the derivative of the distance equation.

Let's denote the position of the particle at time t as P(t) = (x(t), y(t)) = (4cos(t), 2sin(t)).
The distance between P(t) and the point (1, 1) can be calculated using the distance formula:

D(t) = √((x(t) - 1)^2 + (y(t) - 1)^2)

Now, we will differentiate D(t) with respect to t using the chain rule:

dD/dt = dD/dx * dx/dt + dD/dy * dy/dt

First, let's find the individual derivatives:

dD/dx = (x(t) - 1) / √((x(t) - 1)^2 + (y(t) - 1)^2)
dx/dt = -4sin(t)

dD/dy = (y(t) - 1) / √((x(t) - 1)^2 + (y(t) - 1)^2)
dy/dt = 2cos(t)

Substituting these derivatives into the chain rule equation, we have:

dD/dt = ((x(t) - 1)/√((x(t) - 1)^2 + (y(t) - 1)^2)) * (-4sin(t)) + ((y(t) - 1)/√((x(t) - 1)^2 + (y(t) - 1)^2)) * (2cos(t))

Simplifying the equation, we get:

dD/dt = (-4sin(t)(x(t) - 1) + 2cos(t)(y(t) - 1)) / √((x(t) - 1)^2 + (y(t) - 1)^2)

Now, let's find the rate at which the distance is changing when t = π/6:

Substituting t = π/6 into the equation, we have:

dD/dt = (-4sin(π/6)(4cos(π/6) - 1) + 2cos(π/6)(2sin(π/6) - 1)) / √((4cos(π/6) - 1)^2 + (2sin(π/6) - 1)^2)

Simplifying further, we get:

dD/dt = (-4(1/2)(4(√3/2) - 1) + 2(√3/2)(2(1/2) - 1)) / √((4(√3/2) - 1)^2 + (2(1/2) - 1)^2)

Calculating the values:

dD/dt = (-2(√3 - 2) + √3 - 2) / √((2√3 - 1)^2 + (1 - 1)^2)
dD/dt = (-2√3 + 4 + √3 - 2) / √(12 - 4√3 + 3 + 1)
dD/dt = (√3 + 2) / √16 - 4√3 + 4
dD/dt = (√3 + 2) / √(16 - 4√3 + 4)
dD/dt = (√3 + 2) / √(20 - 4√3)

Therefore, the rate at which the distance of the particle to the point (1, 1) is changing when t = π/6 is (√3 + 2) / √(20 - 4√3).

To find the rate at which the distance from the particle to the point (1, 1) is changing at any time t, we need to use the concept of derivatives.

Let's denote the distance between the particle's current position (x, y) and the point (1, 1) as D.

Now, the distance formula between two points (x1, y1) and (x2, y2) is given by:

D = sqrt((x2 - x1)^2 + (y2 - y1)^2)

In our case, (x1, y1) = (1, 1) and (x2, y2) = (x(t), y(t)), where x(t) and y(t) are the x- and y-coordinates of the particle at time t.

So, substituting the values:

D = sqrt((x(t) - 1)^2 + (y(t) - 1)^2)

D = sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2)

Now, we need to find dD/dt, the rate at which the distance D is changing with respect to time t.

To do this, we can take the derivative of D with respect to t using the chain rule:

dD/dt = dD/dx * dx/dt + dD/dy * dy/dt

Let's find the individual derivative terms:

dD/dx = (2 * (4cos(t) - 1)) / (2 * sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2))
dD/dy = (2 * (2sin(t) - 1)) / (2 * sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2))
dx/dt = -4sin(t)
dy/dt = 2cos(t)

Substituting these values into the equation for dD/dt, we get:

dD/dt = [(2 * (4cos(t) - 1)) / (2 * sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2))] * (-4sin(t)) + [(2 * (2sin(t) - 1)) / (2 * sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2))] * (2cos(t))

Simplifying further, we get:

dD/dt = -4sin(t) * (4cos(t) - 1) / sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2) + 2cos(t) * (2sin(t) - 1) / sqrt((4cos(t) - 1)^2 + (2sin(t) - 1)^2)

To find the rate at which the distance is changing when t = pi/6, we can evaluate dD/dt at t = pi/6:

dD/dt = -4sin(pi/6) * (4cos(pi/6) - 1) / sqrt((4cos(pi/6) - 1)^2 + (2sin(pi/6) - 1)^2) + 2cos(pi/6) * (2sin(pi/6) - 1) / sqrt((4cos(pi/6) - 1)^2 + (2sin(pi/6) - 1)^2)

Now we can plug in the values and compute the solution.