The point P on the curve C has parameter p qne the point Q has coordinates (-π, 0). The origin is denoted by O. Given that p is increasing at a constant rate of 0.1 units per second,

Question

find the rate of decrease of triangle OPQ when p=3π/4

parametric equations:

x = θcosθ, y=θsinθ, for 0<θ<π

Answer 1

Let us first draw a picture:

(x=θcosθ, y=θsinθ)  ^  y

for 0<θ<π ___*P__   |

    /     / ` \  \    |    ___

. . .___/  /  `  \  \   |   /    \   /\ C

            /    `   \  \_|_/      \_/  \___. . .

          /      `    \ |      (etc. just an example)

        /       `   \ |            

(-π,0) /       `     \ |(0,0)

<-----Q*----------*R------*O----------------->

The rate of decrease of triangle OPQ when p=3π/4 is the

following, if I assume that you are talking about the area:

Area of OPQ=A(θ)=(1/2)*b*h=(1/2)*(length of line OQ)*(length of line PR)

A(θ)=(1/2)*(π)*(y-coordinate of point P)=πθsinθ/2

Therefore, the rate of change in area of OPQ for all 0<θ<π is:

A'(θ)=dy/dθ=d(πθsinθ/2)/dθ=πsinθ/2+πθcosθ/2

Plugging in θ=p=3π/4 gives us the following:

A'(3π/4) = πsin(3π/4)/2 + π(3π/4)cos(3π/4)/2

A'(3π/4) = π(1/√2)/2 + π(3π/4)(-1/√2)/2 = π/(2√2) - 3(π^2)/(8√2)

A'(3π/4) = 4*[π/(8√2)] - 3π*[π/(8√2)] = -(3π-4)*[π/(8√2)].

So, finally, the rate of decrease (in area) of triangle OPQ is (3π-4)*[π/(8√2)].