A calculus problem by Rocco Dalto

Calculus Level 5

If θ {\bf \theta } is the angle between the two curves 16 x 2 − 24 x y + 9 y 2 − 15 x − 20 y = 0 {\bf 16 x^2 - 24 xy + 9 y^2 - 15 x - 20 y = 0 } and x 2 + y 2 = 2. {\bf x^2 + y^2 = 2.}

Find the positive value for t a n θ . {\bf tan\theta. }


The answer is 3.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Oct 30, 2016

Using the equations of rotation to rotate the x and y axis we have:

x = x ′ c o s θ − y ′ s i n θ {\bf x = x' cos\theta - y' sin\theta }

y = x ′ s i n θ + y ′ c o s θ {\bf y = x' sin\theta + y' cos\theta }

Replacing the equations of rotation in the original equation and simplifying we obtain:

( 16 c o s 2 θ − 12 s i n ( 2 θ ) + 9 s i n 2 θ ) x ′ 2 − ( 7 s i n ( 2 θ ) + 24 c o s ( 2 θ ) ) x ′ y ′ + {\bf (16 cos^2\theta - 12 sin(2\theta) + 9 sin^2\theta)x'^2 - (7 sin(2\theta) + 24 cos(2\theta))x'y' +}

( 16 s i n 2 θ + 12 s i n ( 2 θ ) + 9 c o s 2 θ ) y ′ 2 − 5 ( 3 c o s θ + 4 s i n θ ) x ′ + 5 ( 3 s i n θ − 4 c o s θ ) y ′ = 0. {\bf (16 sin^2\theta + 12 sin(2\theta) + 9 cos^2\theta)y'^2 - 5(3 cos\theta + 4 sin\theta)x' + 5 (3 sin\theta - 4 cos\theta)y' = 0. }

To eliminate the x ′ y ′ {\bf x'y' } term set 7 s i n ( 2 θ ) + 24 c o s ( 2 θ ) = 0 ⟹ t a n ( 2 θ ) = − 24 7 {\bf 7 sin(2\theta) + 24 cos(2\theta) = 0 \implies tan(2\theta) = -\frac{24}{7} }

⟹ t a n 2 θ = 2 t a n θ 1 − t a n 2 θ = − 24 7 ⟹ {\bf \implies tan2\theta = \frac{2 tan\theta}{1 - tan^2\theta} = -\frac{24}{7} \implies }

12 t a n 2 θ − 7 t a n θ − 12 = 0 ⟹ t a n θ = 7 ± 25 24 {\bf 12 tan^2\theta - 7 tan\theta - 12 = 0 \implies tan\theta = \frac{7 \:\pm \:25}{24} }

Since the sign of xy term is negative we choose t a n θ = − 3 4 ⟹ c o s θ = 4 5 {\bf tan\theta = -\frac{3}{4} \implies cos\theta = \frac{4}{\sqrt{5}} } and s i n θ = − 3 5 ⟹ {\bf sin\theta = -\frac{3}{\sqrt{5}} \implies }

x = 4 5 x ′ + 3 5 y ′ {\bf x = \frac{4}{5} x' + \frac{3}{5} y'}

y = − 3 5 x ′ + 4 5 y ′ {\bf y = -\frac{3}{5} x' + \frac{4}{5} y' }

Replacing the equations of rotation in the original equation and simplifying we obtain:

y ′ = x ′ 2 {\bf y' = x'^2 }

So, we have a parabola in the x ′ y ′ {\bf x' y' } system.

The circle x ′ 2 + y ′ 2 = 2 {\bf x'^2 + y'^2 = 2 } is invariant under any rotation

∴ {\bf \therefore } we can write: x ′ 2 + y ′ 2 = 2. {\bf x'^2 + y'^2 = 2 .}

y ′ = x ′ 2 a n d x ′ 2 + y ′ 2 = 2 ⟹ x ′ 4 + x ′ 2 − 2 = 0 ⟹ x ′ 2 = 1 , − 2 {\bf y' = x'^2 \: and \: x'^2 + y'^2 = 2 \implies x'^4 + x'2 - 2 = 0 \implies x'^2 = 1,-2 } and for real x ′ ⟹ x ′ = + − 1 ⟹ {\bf x' \implies x' = +-1 \implies } curves intersect in ( x ′ , y ′ ) {\bf (x',y') } system at ( 1 , 1 ) a n d ( − 1 , 1 ) {\bf (1,1) \: and \: (-1,1)}

For y ′ = x ′ 2 ⟹ d y ′ d x ′ ∣ x ′ = 1 = 2 ∗ x ′ ∣ x ′ = 1 = 2 {\bf y' = x'^2 \implies \frac{dy'}{dx'}|_{x' = 1} = 2 * x'|_{x' = 1} = 2 }

For : x ′ 2 + y ′ 2 = 2 ⟹ d y ′ d x ′ ∣ ( x ′ , y ′ ) = ( 1 , 1 ) = − x y ∣ ( x ′ y ′ ) = ( 1 , 1 ) = − 1 {\bf x'^2 + y'^2 = 2 \implies \frac{dy'}{dx'}|_{(x',y') = (1,1)} = \frac{-x}{y}|_{(x'y') = (1,1)} = -1 }

∴ a t ( x ′ y ′ ) = ( 1 , 1 ) ⟹ t a n θ = − 1 − 2 1 + ( − 1 ) ( 2 ) = 3 , f o r t a n θ > 0. {\bf \therefore \: at \: (x'y') = (1,1) \implies tan\theta = \frac{-1 - 2}{1 + (-1)(2)} = 3, \: for \: tan\theta > 0. }

Similarly, at ( x ′ , y ′ ) = ( − 1 , 1 ) ⟹ t a n θ = 3 , f o r t a n θ > 0. {\bf (x',y') = (-1,1) \implies tan\theta =3, \: for \: tan\theta > 0.} .

Note: I used the graphing calculator to graph the functions and the tangents at the point of intersection of the two functions. I would have liked to put the angle 0 < θ < π 2 {\bf 0 < \theta < \frac{\pi}{2} } in, but I'm not certain how to do it. I have some free programs such as Microsoft Visual Studio 2010 and Free Pascal, and I can write a program in Microsoft Visual Studio 2010 to graph the functions and insert the angle above in, but I'm fairly certain there is a simpler way to do it. Does someone know how to do it? Can you let me know if you do?

1 Helpful 0 Interesting 0 Brilliant 0 Confused

After setting x = x ′ cos ⁡ θ − y ′ sin ⁡ θ x = x' \cos \theta - y' \sin \theta and y = x ′ sin ⁡ θ + y ′ cos ⁡ θ y = x' \sin \theta + y' \cos \theta and substituting into 9 x 2 − 6 x y + y 2 − 12 10 x − 36 10 y = 0 9x^2 - 6xy + y^2 - 12 \sqrt{10} x - 36 \sqrt{10} y = 0 , I get ( 9 cos ⁡ 2 θ − 3 sin ⁡ 2 θ + sin ⁡ 2 θ ) ( x ′ ) 2 − ( 6 cos ⁡ 2 θ + 8 sin ⁡ 2 θ ) ( x ′ y ′ ) + ( cos ⁡ 2 θ + 3 sin ⁡ 2 θ + 9 sin ⁡ 2 θ ) ( y ′ ) 2 + ( − 12 10 cos ⁡ θ − 36 10 sin ⁡ θ ) x ′ + ( 36 10 cos ⁡ θ + 12 10 sin ⁡ θ ) y ′ = 0. \begin{aligned} & (9 \cos^2 \theta - 3 \sin 2 \theta + \sin^2 \theta) (x')^2 - (6 \cos 2 \theta + 8 \sin 2 \theta) (x'y') + (\cos^2 \theta + 3 \sin 2 \theta + 9 \sin^2 \theta) (y')^2 \\ &\quad + (-12 \sqrt{10} \cos \theta - 36 \sqrt{10} \sin \theta) x' + (36 \sqrt{10} \cos \theta + 12 \sqrt{10} \sin \theta) y' = 0. \end{aligned}

I also find your calculations dubious because they do not contain any 10 \sqrt{10} terms.

Jon Haussmann - 4 years, 7 months ago

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The equation should be: 16 x 2 − 24 x y + 9 y 2 − 15 x − 20 y = 0 {\bf 16 x^2 - 24 xy + 9 y^2 - 15 x - 20 y = 0 }

Using the above equation {taken from my solution above) we obtain:

x = 4 5 x ′ + 3 5 y ′ {\bf x = \frac{4}{5} x' + \frac{3}{5} y'}

y = − 3 5 x ′ + 4 5 y ′ {\bf y = -\frac{3}{5} x' + \frac{4}{5} y' }

Replacing the equations of rotation in the original equation and simplifying we obtain:

y ′ = x ′ 2 {\bf y' = x'^2 }

Using 9 x 2 − 6 x y + y 2 − 12 10 x − 36 10 y = 0 {\bf 9x^2 - 6xy + y^2 - 12\sqrt{10}\:x - 36\sqrt{10}\:y = 0 } we would obtain:

x = 1 10 x ′ − 3 10 y ′ {\bf x = \frac{1}{\sqrt{10}} x' - \frac{3}{\sqrt{10}} y'}

y = 3 10 x ′ + 1 10 y ′ {\bf y = \frac{3}{\sqrt{10}} x' + \frac{1}{\sqrt{10}} y' }

Replacing the equations of rotation in the original equation and simplifying we obtain:

⟹ y ′ 2 = 12 x ′ {\bf \implies y'2 = 12 x' }

Rocco Dalto - 4 years, 7 months ago

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