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MrMuchimi
3 years ago
9

How did astronomer Yi Xing (A.D. 683-727) contribute to the development of mathematics in china?

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

Yi Xing invented the astronomical clock and introduced some new methods of interpolation in mathematics.

Step-by-step explanation:

Yi Xing was both an astronomer and a mathematician during the era. He invented the astronomical clock which was more accurate than the initial water and Sun's clock in use.

Furthermore, Yi Xing also discovered some new methods of interpolation in mathematics of which the meaning and interpretation became controversial. Interpolation is a method majorly in mathematics that can be used to estimate a value of a function from its discrete values. It involves first order differences and second order differences.

Also, Yi Xing was able to design a calendar in A.D. 727.

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Read 2 more answers
Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part
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a) The limit of the position of particle Q when time approaches 2 is -\pi.

b) The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2.

c) The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}.

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for t = 2, we need to apply the following two <em>algebraic</em> substitutions:

u = \pi t (1)

k = 2\pi - u (2)

Then, the limit is written as follows:

x(t) =  \lim_{t \to 2} \frac{\sin \pi t}{2-t}

x(t) =  \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}

x(u) =  \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}

x(k) =  \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}

x(k) =  -\pi\cdot  \lim_{k \to 0} \frac{\sin k}{k}

x(k) = -\pi

The limit of the position of particle Q when time approaches 2 is -\pi. \blacksquare

b) The function velocity of particle Q is determined by the <em>derivative</em> formula for the division between two functions, that is:

v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}} (3)

Where:

  • f(t) - Function numerator.
  • g(t) - Function denominator.
  • f'(t) - First derivative of the function numerator.
  • g'(x) - First derivative of the function denominator.

If we know that f(t) = \sin \pi t, g(t) = 2 - t, f'(t) = \pi \cdot \cos \pi t and g'(x) = -1, then the function velocity of the particle is:

v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}

v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}

The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2. \blacksquare

c) The vector <em>rate of change</em> of the distance between particle P and particle Q (\dot r_{Q/P} (t)) is equal to the <em>vectorial</em> difference between respective vectors <em>velocity</em>:

\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t) (4)

Where \vec v_{P}(t) is the vector <em>velocity</em> of particle P.

If we know that \vec v_{P}(t) = (0, 4), \vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right) and t = \frac{1}{2}, then the vector rate of change of the distance between the two particles:

\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)

\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)

\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)

The magnitude of the vector <em>rate of change</em> is determined by Pythagorean theorem:

|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}

|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}

The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}. \blacksquare

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

<em>Particle </em>P<em> moves along the y-axis so that its position at time </em>t<em> is given by </em>y(t) = 4\cdot t - 23<em> for all times </em>t<em>. A second particle, </em>Q<em>, moves along the x-axis so that its position at time </em>t<em> is given by </em>x(t) = \frac{\sin \pi t}{2-t}<em> for all times </em>t \ne 2<em>. </em>

<em />

<em>a)</em><em> As times approaches 2, what is the limit of the position of particle </em>Q?<em> Show the work that leads to your answer. </em>

<em />

<em>b) </em><em>Show that the velocity of particle </em>Q<em> is given by </em>v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}<em>.</em>

<em />

<em>c)</em><em> Find the rate of change of the distance between particle </em>P<em> and particle </em>Q<em> at time </em>t = \frac{1}{2}<em>. Show the work that leads to your answer.</em>

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

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