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Read Symbol defintion
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Q01 |
- Unit circle
- Q02 | - Curve of x = cos(t) and y = sin(t)
- Q03 | - Prove the curve is concave upward when t = 180 to t = 270
- Q04 | - Diagrams of unit circle and unit hyperbola
- Q05 | - Comparison by given diagrams
- Q06 | - Comparison by trigonometric functions
- Q02 | - Curve of x = cos(t) and y = sin(t)
Q01. Unit hyperbola
Pythagorean relation
- sin(t)^2 + cos(t)^2 = 1
- If x = cos(t) and y = sin(t)
- Then x^2 + y^2 = 1
- This is a hyperbola with center at (0,0) and radius r = 0
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Q02. Curve of x = cos(t) and y = sin(t)
Describe the curve when t is from 180 to 270
- From t = 180 to 270
- x = cos(t) = (-)
- y = sin(t) = (-)
- Hence the curve is in 3rd quadrant
- When t = 180 or t = 270
- x = cos(180) = -1
- y = sin(180) = 0
- Hence the point is (-1, 0)
- x = cos(270) = 0
- y = sin(270) = -1
- The curve is from (-1, 0) decreasing to (0, -1)
- The curve is concave upward as shown in diagram
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Q03. Prove the curve is concave upward when t = 180 to t = 270
Proof
- Find first derivative
- y' = dy/dx
- y' = (dy/dt)/(dx/dt)
- Since dy/dt = d/dt(sin(t)) = +cos(t)
- dx/dt = d/dt(cos(t)) = -sin(t)
- Hence y' = (-sin(t))/(+cos(t))
- Hence y' = -tan(t)
- Since t is betwwn 180 and 270, hence y' negative
- Hence the curve is decreasing
- Find second derivative
- y" = d/dt(y')/(dx/dt)
- d/dt(y') = d/dt(-tan(t)) = -sec(t)^2
- Hence y" = -(-sec(t)^2))/(-sin(t))
- Hence y" = -(sec(t)^2)/sin(t)
- Since t is from 180 to 270, sin(t) and sec(t) both negatice
- Hence y" is positive and the curve is concave upward
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Q04 Diagrams of unit circle and hyperbola
- 1. Find the curve of x = cos(t) and y = sin(t)
- 2. Find the curve of x = sin(t) and y = cos(t)
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Q05 Comparison by diagram
Use the diagrams to compare the following two parametric equations
- 1. Equation 1 : x = cos(t) and y = sin(t)
- 2. Equation 2 : x = sin(t) and y = cos(t)
- Based on the diagrams both are circles with center (0,0) and radius = 1
- Henc no difference can be telled
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Q06 Comparison by trigonometric functions
Study the differencce of the following two parametric equations
- 1. Equation 1 : x = cos(t) and y = sin(t)
- 2. Equation 2 : x = sin(t) and y = cos(t)
- x = cos(t) and y = sin(t)
- t = 090 : x = 0 and y = 1
- t = 180 : x = -1 and y = 0
- Hence the arc of circle is from (0, 1) to (-1, 0)
- The curve is concave downward
- The curve in xy coordinate : x from -1 to 0 and y from 0 to 1
- Hence the curve is increasing
- The curve is in 2nd quadrant
- x = sin(t) and y = cos(t)
- t = 090 : x = 1 and y = 0
- t = 180 : x = 0 and y = -1
- Hence the arc of circle is from (1, 0) to (0, -1)
- The curve is concave upward
- The curve in xy coordinate : x from 0 to 1 and y from -1 to 0
- Hence the curve is increasing
- The curve is in 4th quadrant
- Henc difference can be telled
- One is in 2nd quadrant and other in 4th quadrant
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