Inner angle of two functions
- Thread starter tremor
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mmm4444bot
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I'm not sure what "inner angle of two functions" means.
Function f's graph is a parabola, and function g's graph is a line.
The line intersects the parabola at two locations.
Are you talking about the angles formed at each intersection point, where the graph of function g forms one ray and the tangent line to the parabola at the intersection point forms the other ray ?
In this interpretation, the intersection points would each be an angle vertex.
Hello, tremor!
They are referring to the angle formed by the tangents at their points of intersection.
\(\displaystyle \text{The slopes of the tangents are given by the }derivatives\text{ of the functions: }\;\begin{Bmatrix}f'(x) &=& 2x+2 \\ g'(x) &=& 1 \end{Bmatrix}\)
\(\displaystyle \text{The angle }\theta\text{ between two slopes is given by: }\;\tan\theta \;=\;\frac{m_2 - m_1}{1 + m_1m_2}\)
\(\displaystyle \text{Find their intersections: }\;x^2 \,+\, 2x \:=\:x \,+\, 1 \quad\Rightarrow\quad x^2 \,+\, x \,-\, 1 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{-1\pm\sqrt{5}}{2}\)
\(\displaystyle \text{The graphs intersect at }\;\begin{Bmatrix}\text{point }P: & x \:=\:\dfrac{-1+\sqrt{5}}{2} \\ \text{point }Q: &x \:=\:\dfrac{-1-\sqrt{5}}{2} \end{Bmatrix}\)
\(\displaystyle \text{At point }P:\; \begin{array}{ccccccccc}f'\left(\frac{-1+\sqrt{5}}{2}\right) &=& 2\left(\frac{-1+\sqrt{5}}{2}\right)+2 &=& 1 + \sqrt{5} &=& m_2 \\ g'\left(\frac{-1+\sqrt{5}}{2}\right) &&& =& 1 &=& m_1 \end{array}\)
. . \(\displaystyle \tan\theta_1 \;=\;\frac{(1+\sqrt{5}) - 1}{1 + (1+\sqrt{5})\cdot1} \;=\;\frac{\sqrt{5}}{2 + \sqrt{5}} \;=\; 0.527864045 \quad\Rightarrow\quad \boxed{\theta_1 \;\approx\;27.8^o}\)
\(\displaystyle \text{At point }Q:\; \begin{array}{ccccccccc}f'\left(\frac{-1-\sqrt{5}}{2}\right) &=& 2\left(\frac{-1-\sqrt{5}}{2}\right)+2 &=& 1 - \sqrt{5} &=& m_1 \\ g'\left(\frac{-1-\sqrt{5}}{2}\right) &&& =& 1 &=& m_2 \end{array}\)
. . \(\displaystyle \tan\theta_2 \;=\;\frac{(1 - (1-\sqrt{5})}{1 + 1\cdot(1-\sqrt{5})} \;=\;\frac{-\sqrt{5}}{2 - \sqrt{5}} \;=\; 9.472135955 \quad\Rightarrow\quad \boxed{\theta_2 \;\approx\;84.0^o}\)
They are referring to the angle formed by the tangents at their points of intersection.
\(\displaystyle \text{The slopes of the tangents are given by the }derivatives\text{ of the functions: }\;\begin{Bmatrix}f'(x) &=& 2x+2 \\ g'(x) &=& 1 \end{Bmatrix}\)
\(\displaystyle \text{The angle }\theta\text{ between two slopes is given by: }\;\tan\theta \;=\;\frac{m_2 - m_1}{1 + m_1m_2}\)
\(\displaystyle \text{Find their intersections: }\;x^2 \,+\, 2x \:=\:x \,+\, 1 \quad\Rightarrow\quad x^2 \,+\, x \,-\, 1 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{-1\pm\sqrt{5}}{2}\)
\(\displaystyle \text{The graphs intersect at }\;\begin{Bmatrix}\text{point }P: & x \:=\:\dfrac{-1+\sqrt{5}}{2} \\ \text{point }Q: &x \:=\:\dfrac{-1-\sqrt{5}}{2} \end{Bmatrix}\)
\(\displaystyle \text{At point }P:\; \begin{array}{ccccccccc}f'\left(\frac{-1+\sqrt{5}}{2}\right) &=& 2\left(\frac{-1+\sqrt{5}}{2}\right)+2 &=& 1 + \sqrt{5} &=& m_2 \\ g'\left(\frac{-1+\sqrt{5}}{2}\right) &&& =& 1 &=& m_1 \end{array}\)
. . \(\displaystyle \tan\theta_1 \;=\;\frac{(1+\sqrt{5}) - 1}{1 + (1+\sqrt{5})\cdot1} \;=\;\frac{\sqrt{5}}{2 + \sqrt{5}} \;=\; 0.527864045 \quad\Rightarrow\quad \boxed{\theta_1 \;\approx\;27.8^o}\)
\(\displaystyle \text{At point }Q:\; \begin{array}{ccccccccc}f'\left(\frac{-1-\sqrt{5}}{2}\right) &=& 2\left(\frac{-1-\sqrt{5}}{2}\right)+2 &=& 1 - \sqrt{5} &=& m_1 \\ g'\left(\frac{-1-\sqrt{5}}{2}\right) &&& =& 1 &=& m_2 \end{array}\)
. . \(\displaystyle \tan\theta_2 \;=\;\frac{(1 - (1-\sqrt{5})}{1 + 1\cdot(1-\sqrt{5})} \;=\;\frac{-\sqrt{5}}{2 - \sqrt{5}} \;=\; 9.472135955 \quad\Rightarrow\quad \boxed{\theta_2 \;\approx\;84.0^o}\)
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* |so the original problem is stated like this:
compute the inner products and angles between the following functions using the definition for inner product as:
\(\displaystyle <f(t),g(t)>=\int_0^1 \! f(t)g(t) \, \mathrm{d}t\)
i have calculated the inner products from this but i feel badly i hadn't given you all the information in the problem. is it still the tangents as you described that you think they're looking for?
compute the inner products and angles between the following functions using the definition for inner product as:
\(\displaystyle <f(t),g(t)>=\int_0^1 \! f(t)g(t) \, \mathrm{d}t\)
i have calculated the inner products from this but i feel badly i hadn't given you all the information in the problem. is it still the tangents as you described that you think they're looking for?
The angle can be found by using \(\displaystyle cos{\theta}=\frac{f\cdot g}{||f||\cdot ||g||}\)
Where \(\displaystyle \int_{0}^{1}f(t)g(t)dt\) is the dot product. If it is 0, then they are orthogonal.
The norms can be found by using \(\displaystyle \sqrt{\int_{0}^{1}f^{2}(t)dt}\)
It has been awhile since I used this stuff.
Try googling orthogonal, inner products, angles between functions.
The thing is, this gives me 22.265 degrees. I like Soroban's method. I came up with the same result that way.
But,
\(\displaystyle \frac{\int_{0}^{1}[(t^{2}+2t)(t+1)]dt}{\sqrt{\int_{0}^{1}(t^{2}+2t)^{2}dt}\cdot \sqrt{\int_{0}^{1}(t+1)^{2}dt}}=\frac{\frac{9}{4}}{\frac{\sqrt{570}}{15}\cdot \frac{\sqrt{21}}{3}}\)
\(\displaystyle cos^{-1}\left(\frac{\frac{9}{4}}{\frac{\sqrt{570}}{15}\cdot \frac{\sqrt{21}}{3}}\right)=22.265^{o}\)
Where \(\displaystyle \int_{0}^{1}f(t)g(t)dt\) is the dot product. If it is 0, then they are orthogonal.
The norms can be found by using \(\displaystyle \sqrt{\int_{0}^{1}f^{2}(t)dt}\)
It has been awhile since I used this stuff.
Try googling orthogonal, inner products, angles between functions.
The thing is, this gives me 22.265 degrees. I like Soroban's method. I came up with the same result that way.
But,
\(\displaystyle \frac{\int_{0}^{1}[(t^{2}+2t)(t+1)]dt}{\sqrt{\int_{0}^{1}(t^{2}+2t)^{2}dt}\cdot \sqrt{\int_{0}^{1}(t+1)^{2}dt}}=\frac{\frac{9}{4}}{\frac{\sqrt{570}}{15}\cdot \frac{\sqrt{21}}{3}}\)
\(\displaystyle cos^{-1}\left(\frac{\frac{9}{4}}{\frac{\sqrt{570}}{15}\cdot \frac{\sqrt{21}}{3}}\right)=22.265^{o}\)
mmm4444bot
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soroban said:
Ah, thank you for that definition.
Assuming that the tangents are not perpendicular, does the name "inner angle" always refer to the smaller of the two angles formed ?
What are you integrating over?. \(\displaystyle (0,1)\)?.
If so, then they are orthogonal.
\(\displaystyle \int_{0}^{1}\left(sin(2\pi t)cos(2\pi t)\right)dt=0\)
Also, since sin is odd and cos even, then the product of them is odd. Therefore, 0.
If so, then they are orthogonal.
\(\displaystyle \int_{0}^{1}\left(sin(2\pi t)cos(2\pi t)\right)dt=0\)
Also, since sin is odd and cos even, then the product of them is odd. Therefore, 0.