Limiting Angle

Calculus Level 4

A ( 1 , 0 ) A(1, 0) is a point on the parabola y = 2 x ( x 1 ) y=2x(x-1) . From point A A , point P P is moving along the curve towards the origin O ( 0 , 0 ) O(0, 0) . What is the limit lim P O sec A P O ? \lim_{P \rightarrow O } \sec \angle APO ?

1 3 -\frac{1}{\sqrt{3}} 5 -\sqrt{5} 1 5 -\frac{1}{\sqrt{5}} 3 -\sqrt{3}

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3 solutions

Anish Puthuraya
Feb 7, 2014

There must be other ways to solve this, but I found this method quite traditional.

First of all,
Letting A P O = α \angle APO = \alpha ,
If sec α \displaystyle\sec\alpha has a limit, then sin α \displaystyle\sin\alpha also has a limit. Thus, we shall first find \displaystyle\sin\alpha).

Consider a general point P ( x , y ) \displaystyle P(x,y) , which lies on the given parabola.
It is clear that,
y = 2 x ( x 1 ) \displaystyle y = 2x(x-1)

Using Sine Rule,
sin α O A = sin A O P A P \displaystyle\frac{\sin\alpha}{OA} = \frac{\sin\angle AOP}{AP}

Thus,
sin α 1 = y x 2 + y 2 A P = y x 2 + y 2 ( x 1 ) 2 + y 2 \displaystyle\frac{\sin\alpha}{1} = \frac{\frac{y}{\sqrt{x^2+y^2}}}{AP} = \frac{\frac{y}{\sqrt{x^2+y^2}}}{\sqrt{(x-1)^2+y^2}}

Substituting y = 2 x ( x 1 ) \displaystyle y=2x(x-1) , we get,
sin α = 2 x ( x 1 ) ( 4 x 4 8 x 3 + 5 x 2 ) ( 4 x 4 8 x 3 + 5 x 2 2 x + 1 ) = 2 ( x 1 ) ( 4 x 2 8 x + 5 ) ( 4 x 4 8 x 3 + 5 x 2 2 x + 1 ) \sin\alpha=\frac{2x(x-1)}{\sqrt{(4x^4-8x^3+5x^2)(4x^4-8x^3+5x^2-2x+1)}}=\frac{2(x-1)}{\sqrt{(4x^2-8x+5)(4x^4-8x^3+5x^2-2x+1)}}

Note that we cancelled out the x \displaystyle x 's, which is why we could not just substitute the final x-coordinates i.e x = 0 \displaystyle x=0

Now, substituting x = 0 \displaystyle x=0 , because, as the problem suggests, finally, x 0 \displaystyle x\to 0

sin α = 2 5 \displaystyle\sin\alpha = \frac{-2}{\sqrt{5}}

cos α = ± 1 5 \displaystyle\Rightarrow\cos\alpha = \pm\frac{1}{\sqrt{5}}

We reject the positive value, because, when P \displaystyle P approaches O \displaystyle O , clearly, A P O > 9 0 o \displaystyle\angle APO > 90^o
sec ( A P O ) < 0 \displaystyle\Rightarrow\sec(APO)<0

Thus,
sec α = 5 \displaystyle\sec\alpha = -\sqrt{5}

We can also use vectors to find the angle between vectors PA and PO.

Karthik Kannan - 7 years, 4 months ago

The slope at the origin is 2 -2 (I used calculus to get that so I guess I sorta cheated). Let angle A O P = θ AOP = \theta . Then

tan θ = 2 \tan \theta = -2 . In this case θ \theta is negative but we will take the positive value of theta for our purposes.

Note that as point P P approaches point O O P A O \angle PAO approaches 0 0 . This means that the limiting value of A P O = 180 θ \angle APO = 180 - \theta , call it α \alpha . We may realize that

tan α = 1 2 \tan \alpha = \frac{-1}{2} .

Now we may construct a right triangle with sides 2 2 and 1 1 and hypotenuse equal to 5 \sqrt{5} (by the Pythagorean Theorem ), and from here it is simple to deduce that sec α = 1 5 \sec \alpha = \frac{-1}{\sqrt{5}} .

Saud Molaib - 7 years, 4 months ago
Julio Reyes
Feb 11, 2014

Recall: The slope m ( P O ) of the secant line P O is the average rate of change between P and O lim P O m ( P O ) = Slope of the tangent line at point O \text{Recall: } \\ \text{The slope } m(\overrightarrow{PO}) \text{ of the secant line } \overrightarrow{PO} \text{ is the average rate of change between } P \text{ and } O \\ \lim_{_{P \to O}} m(\overrightarrow{PO}) = \text{Slope of the tangent line at point } O \\

What the problem is essentially asking is the secant of the angle between the tangent line at point O and the line O A . \text{What the problem is essentially asking is the secant of the angle between } \\ \text{the tangent line at point O and the line } \overrightarrow{OA}.

http://postimg.org/image/ssy8jb1ll/ http://postimg.org/image/ssy8jb1ll/

To solve the problem we need to find the angle θ using the dot product of T and O A . To do this we must first find vector T . \mbox{ } \\ \text{To solve the problem we need to find the angle } \theta \text{ using the dot product of } \mathbf{T} \text { and } \mathbf{OA}. \\ \text{To do this we must first find vector } \mathbf{T}.

m ( x ) = d d x ( 2 x ( x 1 ) ) = 4 x 2 and m ( 0 ) = 2 T = 1 , 2 and O A = 1 , 0 Note: To find vector T just choose two points on the line y = -2x, e.g (-1,2) and (0,0), and subtract their components. m(x) = \frac{\mathrm d}{\mathrm d x} \big( 2x(x -1) \big) = 4x - 2 \quad \text{and} \quad m(0) = -2 \\ \mathbf{T} = \langle -1, 2 \rangle \quad \text{and} \quad \mathbf{OA} = \langle 1, 0 \rangle \\ _{\text{Note: To find vector T just choose two points on the line y = -2x, e.g (-1,2) and (0,0), and subtract their components.}} \\

Now that we have the two vectors we can solve the problem. \text{Now that we have the two vectors we can solve the problem. }

T O A = 1 T = 5 O A = 1 T O A = T O A c o s ( θ ) c o s ( θ ) = T O A T O A = 1 5 \mathbf{T}\bullet \mathbf{OA} = -1 \quad \quad | \mathbf{T} | = \sqrt{5} \quad \quad | \mathbf{OA} | = 1 \\ \mathbf{T}\bullet \mathbf{OA} = | \mathbf{T} | \mbox{ } | \mathbf{OA } | \mbox{ } cos(\theta ) \quad \rightarrow \quad cos(\theta) = {{\mathbf{T} \bullet \mathbf{OA}} \over {| \mathbf{T} | | \mathbf{OA}|}} = {-1 \over \sqrt{5}} \\

Finally: s e c ( θ ) = 1 c o s ( θ ) = 5 \text{Finally:} \\ sec(\theta) = {1 \over cos(\theta)} = \boxed{- \sqrt{5}}

Siladitya Basu
Jul 3, 2014

My solution involves finding A P O \angle APO in terms of the co-ordinates of point P P and then taking the limit as P O P \to O .

Let P ( t , 2 t ( t 1 ) ) P \equiv (t, 2t(t-1)) , as it lies on the parabola y = 2 x ( x 1 ) y=2x(x-1) .

Slope of O P , m 1 = 2 t ( t 1 ) 0 t 0 = 2 ( t 1 ) OP, m_1 = \frac{2t(t-1)-0}{t-0} = 2(t-1) .

Slope of P A , m 2 = 2 t ( t 1 ) 0 t 1 = 2 t PA, m_2 = \frac{2t(t-1)-0}{t-1} = 2t .

Let A P O = θ \angle APO = \theta , then tan θ = m 2 m 1 1 + m 1 m 2 \left|\tan \theta \right|= \left| \frac{m_2-m_1}{1+m_1m_2}\right| .

Since sec 2 θ = 1 + tan 2 θ \sec^2 \theta = 1 + \tan^2 \theta , we have sec θ = ± 1 + ( 2 1 + 4 t ( t 1 ) ) 2 \sec \theta = \pm \sqrt{1 + \left(\frac{2}{1+4t(t-1)}\right)^2} .

Observe that as P O , t 0 P \to O, t \to 0 ; and also A P O > 9 0 \angle APO > 90^\circ , so sec A P O < 0 \sec \angle APO < 0 .

lim P O sec A P O = lim t 0 1 + ( 2 1 + 4 t ( t 1 ) ) 2 = 5 \lim_{P \to O} \sec \angle APO = - \lim_{t \to 0} \sqrt{1 + \left(\frac{2}{1+4t(t-1)}\right)^2} = -\sqrt{5} .

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