Date | May 2019 | Marks available | 2 | Reference code | 19M.2.AHL.TZ1.H_10 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Find | Question number | H_10 | Adapted from | N/A |
Question
The voltage v in a circuit is given by the equation
v(t)=3sin(100πt), t⩾0 where t is measured in seconds.
The current i in this circuit is given by the equation
i(t)=2sin(100π(t+0.003)).
The power p in this circuit is given by p(t)=v(t)×i(t).
The average power pav in this circuit from t=0 to t=T is given by the equation
pav(T)=1T∫T0p(t)dt, where T>0.
Write down the maximum and minimum value of v.
Write down two transformations that will transform the graph of y=v(t) onto the graph of y=i(t).
Sketch the graph of y=p(t) for 0 ≤ t ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.
Find the total time in the interval 0 ≤ t ≤ 0.02 for which p(t) ≥ 3.
Find pav(0.007).
With reference to your graph of y=p(t) explain why pav(T) > 0 for all T > 0.
Given that p(t) can be written as p(t)=asin(b(t−c))+d where a, b, c, d > 0, use your graph to find the values of a, b, c and d.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3, −3 A1A1
[2 marks]
stretch parallel to the y-axis (with x-axis invariant), scale factor 23 A1
translation of (−0.0030) (shift to the left by 0.003) A1
Note: Can be done in either order.
[2 marks]
correct shape over correct domain with correct endpoints A1
first maximum at (0.0035, 4.76) A1
first minimum at (0.0085, −1.24) A1
[3 marks]
p ≥ 3 between t = 0.0016762 and 0.0053238 and t = 0.011676 and 0.015324 (M1)(A1)
Note: Award M1A1 for either interval.
= 0.00730 A1
[3 marks]
pav=10.007∫0.00706sin(100πt)sin(100π(t+0.003))dt (M1)
= 2.87 A1
[2 marks]
in each cycle the area under the t axis is smaller than area above the t axis R1
the curve begins with the positive part of the cycle R1
[2 marks]
a=4.76−(−1.24)2 (M1)
a=3.00 A1
d=4.76+(−1.24)2
d=1.76 A1
b=2π0.01
b=628(=200π) A1
c=0.0035−0.014 (M1)
c=0.00100 A1
[6 marks]