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Q01 |
- Diagram : Graph of x = t^3 - 3*t and y = t^2 + 1
- Q02 | - Sketch the curve
- Q03 | - Find intersection of the curve
- Q04 | - Find the angle at intersection
- Q02 | - Sketch the curve
Q01. Diagram : Curve of x = t^3 - 3*t and y = t^2 + 1
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Q02. Sketch the curve
Data points of x = t^3 - 3*t and y = t^2 + 1
- t .... 0 .... 1 ... Sqr(3) .... 2 ....
- x .... 0 ... -2 ........ 0 .... 2 ....
- y .... 1 .... 2 ........ 4 .... 5 ....
- t .... 0 ... -1 ... -Sqr(3) .... -2 ....
- x .... 0 ... +2 ........ 0 ..... -2 ....
- y .... 1 .... 2 ........ 4 ...... 5 ....
- 1. The curve is symmetrical to y-axis
- 2. The curve is intecepted at (0, 4)
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Q03. Find intersection of the curve
Answer
- From Q02, we know the intersection is at (0, 4)
- We can also see the intersection from the diagram
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Q04. Find angle at intersection of the curve
Find the first derivative of x = t^3 - 3*t and y = t^2 + 1
- dy/dx = (dy/dt)/(dx/dt)
- dy/dx = (3*t^2 - 3)/(2*t)
- Slope at point (0, 4) when t = Sqr(3)
- The value of t is Sqr(3)
- Hence slope is (3*3 - 3)/(2*Sqr(3)) = 6/3.4641 = 1.7320516
- Slope = tan(A1)
- Hence A1 = arctan(1.7320516) = 60 degrees
- Slope at point (0, 4) when t = -Sqr(3)
- The value of t is -Sqr(3)
- Hence slope is (3*3 - 3)/(-2*Sqr(3)) = -6/3.4641 = -1.7320516
- Slope = tan(A1)
- Hence A1 = arctan(-1.7320516) = 120 degrees
- Hence angle at intersection is (120 - 60) = 60
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