help! calculus!?

Question

1. point A moves along a unit circle at the rate of 2 units/second counterclockwise. What is the rate of change of the perimeter of the triangle whose vertices are points A, B(1,0) and C(-1,0) when A is at the point with coordinate (1/2, (sqrt3)/2).

Answer 1

Here's how I did it. If the circle is a unit circle, then B and C make up the diameter (horizontally). Geometry says that if a triangle is inscribed in a circle such that one leg is the diameter, then it's a right triangle. So angle A is always 90 degrees.

Let θ be the angle AOB (where O is the center of the circle). Since A is on the unit circle, the coordinates of A are simply (cosθ, sinθ). Using the distance formula, this means AB is
√[(cosθ - 1)² + (sinθ - 0)²] = √[(cos² θ - 2cosθ + 1) + sin² θ] =
√[(cos² θ + sin² θ) + 1 - 2cosθ)] = √[2-2cos(θ)]. Using Pythagoras, AC² + AB² = 2², so AC = √[4 - [2-2cos(θ)]] =
√[2+2cos(θ)]. The perimeter then is P(θ) = 2 + √[2+2cos(θ)] + √[2-2cos(θ)]

If the point moves along the circumference (which is π*1 = π) at 2 units/second, then it travels the whole circle in (1 second/2 units)*(π) = π/2 seconds. Its change in angle per second is the sweep of the entire circle (2π) divided by the number of seconds (π/2), or 4 rad/s.

To get the rate of change for the perimeter, I suppose you could get dP/dt by taking dP/dθ, writing in "dθ/dt" when you'd expect ot use the chain rule, then subsituting 4 in for the dθ/dt values. Given the coordinates, you should be able to find θ and plug it into the derivative.

Answer 2

I assume a unit is the distance 1. For a unit circle, that is also 1 radian.

Let
p = perimeter of triangle
dθ/dt = 2 units/sec = 2 radians/sec

Given
B(1,0)
C(-1,0)
A(1/2,√3/2) moving at 2 radians/sec counterclockwise

Find dp/dt.

p = 2 + √{(x - 1)² + y²} + √{(x + 1)² + y²}
p = 2 + √{x² - 2x + 1 + y²} + √{(x² + 2x + 1 + y²}
p = 2 + √(2 - 2x) + √(2 + 2x)

But x = cosθ

p = 2 + √(2 - 2x) + √(2 + 2x)
p = 2 + √(2 - 2cosθ) + √(2 + 2cosθ)

dp/dθ = (½)(-2)(-sinθ)/√(2 - 2cosθ) + (½)(2)(-sinθ)/√(2 + 2cosθ)
dp/dθ = sinθ/√(2 - 2cosθ) - sinθ)/√(2 + 2cosθ)
dp/dθ = (√3/2)/√(2 - 2*½) - (√3/2)/√(2 + 2*½)
dp/dθ = (√3/2)/√(2 - 1) - (√3/2)/√(2 + 1)
dp/dθ = (√3/2) - (√3/2)/√3 = √3/2 - 1/2

dp/dt = (dp/dθ)(dθ/dt) = {√3/2 - 1/2}*2 = √3 - 1

Answer 3

A = (cos theta, sin theta), where theta is the angle between the line (origin, A) and the X-axis. You need going to find the derivative of that perimeter with respect to theta, evaluated at theta = pi/3, and then multiply it by a constant to get the rate of change with respect to time. (What constant? Well, it takes pi/2 seconds for theta to increase by pi.)

So do some trigonometry to find the length of each of the non-hypoteneuse sides. The length of the hypoteneuse is of course 2. Add up the lengths of the three sides, and you have the perimeter as a function of theta.

Answer 4

Find an expression for perimeter P in terms of x: (BC=2, I used distance formula for AB, AC)
P= 2 + sqrt((x+1)^2+y^2) + sqrt((x-1)^2+y^2)
then expand under the square roots
P= 2 + sqrt(x^2+2x+1+y^2) + sqrt(x^2-2x+1+y^2)
and since it's a unit circle x^2+y^2=1 you get
P= 2 + sqrt(2+2x) + sqrt(2-2x).

Then differentiate P w.r.t x.
dP/dx = 1/2*2*(2+2x)^(-1/2) -1/2*2*(2-2x)^(-1/2).
= (2+2x)^(-1/2) -(2-2x)^(-1/2).

Then to find rate of change of P (w.r.t time t) use change of variable formula
dP/dt = dP/dx * dx/dt.
You want to know what the value of dP/dt is when x=1/2 so substitute x=1/2 into dP/dx.

dP/dx = [(2+1)^(-1/2)-(2-1)^(-1/2)]
= 3^(-1/2)-1
= 1/sqrt(3) -1.

and then you're told in the question that dx/dt =2 so
dP/dt = dP/dx * dx/dt
= 2(1/sqrt(3) -1).

I think that's right :-S. Hope it helps.

Answer 5

The notation is too difficult in this text format. My solution is here
http://img179.imageshack.us/img179/7272/triangleincirclebo7.png

Please check for any errors, but the process should be clear.

NOTE: I took the "2 units" to be 2 radians. If the circle is a unit circle, then one "unit" of arc subtends one radian.