Sometimes It Is Irresistible To Be Sinusoidal

Calculus Level pending

Consider a circle with radius r r and center O. Let the endpoints of its diameter be A A and B B respectively. Now consider a variable point P P on the circumference of the circle moving in an anticlockwise sense, starting from B B . If ∠ P O B = θ \angle POB=\theta then,

The rate of change of area of Δ APB \Delta\text{APB} with respect to θ \theta when θ = π / 3 \theta=\pi /3 is a r 2 / b ar^2/b , where a a and b b are coprime positive integers.

Find b a + b b^{a+b} .


The answer is 8.

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
Jan 27, 2017

Let the coordinates of P be ( r cos ⁡ θ , r sin ⁡ θ (r\cos\theta, r\sin\theta . Then the area of area of the triangle: f ( θ ) = 1 2 ⋅ ( 2 r ) ⋅ ( r sin ⁡ θ ) = r 2 sin ⁡ θ f(\theta) =\dfrac 12\cdot(2r)\cdot (r\sin\theta) =r^2\sin\theta

Therefore, d f ( θ ) d θ = r 2 cos ⁡ θ \dfrac{df(\theta)}{d\theta}=r^2\cos\theta . And at θ = π / 3 \theta =\pi /3 , d f ( θ ) d θ π / 3 = r 2 cos ⁡ ( π / 3 ) = r 2 2 \dfrac{df(\theta)}{d\theta_{\pi /3}}=r^2\cos(\pi/3)=\frac{r^2} 2

Hence, a = 1 a=1 , b = 2 b=2 and b a + b = 2 3 = 8 b^{a+b}=2^3=8

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