A particle is moving around the ellipse 4x^2 + 16y^2 =64. At any time the x and y coordinates are given by the expressions X(t)=4cost and y(t)=2sint. At what rate is the particle distance from the point (2,0) changing when t = π/4Answer (round to two decimal places)
Question
A particle is moving around the ellipse 4x^2 + 16y^2 =64. At any time the x and y coordinates are given by the expressions X(t)=4cost and y(t)=2sint. At what rate is the particle distance from the point (2,0) changing when t = π/4Answer (round to two decimal places)
Solution
The distance between the particle and the point (2,0) at any time t is given by the formula:
D(t) = sqrt[(X(t) - 2)^2 + (Y(t) - 0)^2]
Substituting the given expressions for X(t) and Y(t), we get:
D(t) = sqrt[(4cos(t) - 2)^2 + (2sin(t))^2]
To find the rate at which this distance is changing, we need to differentiate D(t) with respect to t. This requires the chain rule and the derivatives of the sine and cosine functions.
The derivative of D(t) is:
D'(t) = 1/2 * [2(4cos(t) - 2)(-4sin(t)) + 4(2sin(t))(2cos(t))] / sqrt[(4cos(t) - 2)^2 + (2sin(t))^2]
Simplify this expression to get:
D'(t) = -8(cos(t)sin(t) + sin^2(t)) / sqrt[(4cos(t) - 2)^2 + (2sin(t))^2]
Now substitute t = π/4 into this expression to find the rate at which the distance is changing at that time.
D'(π/4) = -8(cos(π/4)sin(π/4) + sin^2(π/4)) / sqrt[(4cos(π/4) - 2)^2 + (2sin(π/4))^2]
After calculating the above expression, round your answer to two decimal places.
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