Another differential equation problem

Calculus Level 5

We have a continous,differentiable curve passing through ( 0 , 1 ) (0,1) defined as f ( x , y ) = 0 f(x,y)=0 satisfying the property that the area made by tangent and normal to the curve at any point on the curve ( x , y ) (x,y) and the x-axis is equal to the magnitude of the ordinate(i.e. = ∣ y ∣ =\left| y\right| ).

There are exactly two possible curves satisfying the condition. One is bounded and the other one is unbounded curves.

Let the area bounded by the bounded curve and the y-axis be = a − b π c a-\frac{b\pi}{c} ,where a , b , c a,b,c are positive integers and b , c b,c are co-prime.

Find a + b + c a+b+c

Details and assumptions

1) I am saying unbounded curve in the sense that for some x either y is tending to infinity or for some y,x is tending to infinity.


The answer is 5.

Ronak Agarwal by Ronak Agarwal , 23, India

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1 solution

Wesley Low
Jul 30, 2020

Consider a point ( a , b ) \left(a,b\right) with tangent and normal lines ( y − b ) = ( tan ⁡ θ ) ( x − a ) \left(y-b\right)=\left(\tan\theta\right)\left(x-a\right) and ( y − b ) = ( − cot ⁡ θ ) ( x − a ) \left(y-b\right)=\left(-\cot\theta\right)\left(x-a\right) respectively. The base of the triangle defined by these 2 lines and the x-axis has a length b ( tan ⁡ θ + cot ⁡ θ ) b\left(\tan\theta+\cot\theta\right) . For a given triangle where area = height, it's base length must be 2. Hence, b ( tan ⁡ θ + cot ⁡ θ ) = 2 b\left(\tan\theta+\cot\theta\right)=2 b = 2 tan ⁡ θ + cot ⁡ θ b=\frac{2}{\tan\theta+\cot\theta} = 2 sin ⁡ θ cos ⁡ θ =2\sin\theta\cos\theta ( a , b ) = ( a , 2 sin ⁡ θ cos ⁡ θ ) \left(a,b\right)=\left(a,2\sin\theta\cos\theta\right)

Using the Chain Rule, d a d θ = d b d θ d b d a \frac{da}{d\theta}=\frac{\frac{db}{d\theta}}{\frac{db}{da}} a = ∫ 2 ( cos ⁡ 2 θ − sin ⁡ 2 θ ) tan ⁡ θ d θ a=\int\frac{2\left(\cos^{2}\theta-\sin^{2}\theta\right)}{\tan\theta}d\theta = 2 ∫ ( cos ⁡ 3 θ sin ⁡ θ − sin ⁡ θ cos ⁡ θ ) d θ =2\int\left(\frac{\cos^{3}\theta}{\sin\theta}-\sin\theta\cos\theta \right)d\theta a = 2 ( ln ⁡ sin ⁡ θ − sin ⁡ 2 θ ) + C a=2\left(\ln\sin\theta-\sin^{2}\theta\right)+C

Since the point ( 0 , 1 ) \left(0,1\right) lies on the curve, C = ln ⁡ 2 + 1 C=\ln{2}+1 ( a , b ) = ( 2 ( ln ⁡ sin ⁡ θ − sin ⁡ 2 θ ) + ln ⁡ 2 + 1 , 2 sin ⁡ θ cos ⁡ θ ) \left(a,b\right)=\left(2\left(\ln\sin\theta-\sin^{2}\theta\right)+\ln{2}+1,2\sin\theta\cos\theta\right)

The required area is defined along this curve at θ ∈ { [ − π 2 , − π 4 ] ∪ [ π 4 , π 2 ] } \theta\in\left\{\left[-\frac{\pi}{2},-\frac{\pi}{4}\right]\cup\left[\frac{\pi}{4},\frac{\pi}{2}\right]\right\} . Hence the area is ∣ 2 ∫ π 4 π 2 ( 2 sin ⁡ θ cos ⁡ θ ) d ( 2 ( ln ⁡ sin ⁡ θ − sin ⁡ 2 θ ) + ln ⁡ 2 + 1 ) ∣ \left|2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(2\sin\theta\cos\theta\right)d\left(2\left(\ln\sin\theta-\sin^{2}\theta\right)+\ln{2}+1\right)\right| = ∣ 8 ∫ π 4 π 2 ( sin ⁡ θ cos ⁡ θ ) ( cot ⁡ θ − 2 sin ⁡ θ cos ⁡ θ ) d θ ∣ =\left|8\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\sin\theta\cos\theta\right)\left(\cot\theta-2\sin\theta\cos\theta\right)d\theta\right| = 2 − π 2 =\boxed{2-\frac{\pi}{2}}

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