Asked by rhine
find the period, amplitude,zeros, extreme points at maximum and minimum of the given function:
1.y = sin( -2x)
2.y = 4 sin( 5x)
3.y = -3 sin (3x/5)
4.y = 1/2 cos (-4x)
1.y = sin( -2x)
2.y = 4 sin( 5x)
3.y = -3 sin (3x/5)
4.y = 1/2 cos (-4x)
Answers
Answered by
Reiny
your equations all fit the pattern
y = a (sin or cos)(kx)
where |a| is the amplitude and the period is 2π/k
I will do one of them , you do the others in the same way.
2.
y = 4sin(5x)
amplitude is 4
period = 2π/5 radians
dy/dx = 4cos(5x) (5) = 20 cos(5x)
= 0 for any max/min
20cos(5x) = 0
cos(5x) = 0
I know cos(π/2) = 0 and cos(3π/2) = 0
so 5x = π/2 or 5x = 3π/2
x = π/10 or x = 3π/10
when x = π/10 , y = 4sin(5(π/10)) = 4sin(π/2) = 4
when x = 3π/10 y = 4sin(5(3π/10)=4sin(3π/2) = -4
So in the first period, the max point is (π/10 , 4)
and the min point is (3π/10 , -4)
I used Calculus to find the max/min.
We could have done this by simply knowing something about our curve.
We know we have a sine curve of period 2π/5
So mark off a section of the line from 0 to 2π/5
and marking its 4 quadrants as
0 π/10 2π/10 3π/10 and 4π/10
or
0 , π/10 , π/5 , 3π/10 , and 2π/5
We know this sine curve starts at 0 goes up to 4 at π/10, back to 0 at π/5 , down to -4 at 3π/10 and then back up to 0 at 2π/5
you did not state in what domain you want these, so I just did the first period.
Of course each maximum happens again in each consecutive or previous period
so our max's happen at x = π/10 + k(2π/5) , where k is an integer
and our min's happen at x = 3π/10 + k(2π/5)
Of course each max will be 4 and each min will be -4
y = a (sin or cos)(kx)
where |a| is the amplitude and the period is 2π/k
I will do one of them , you do the others in the same way.
2.
y = 4sin(5x)
amplitude is 4
period = 2π/5 radians
dy/dx = 4cos(5x) (5) = 20 cos(5x)
= 0 for any max/min
20cos(5x) = 0
cos(5x) = 0
I know cos(π/2) = 0 and cos(3π/2) = 0
so 5x = π/2 or 5x = 3π/2
x = π/10 or x = 3π/10
when x = π/10 , y = 4sin(5(π/10)) = 4sin(π/2) = 4
when x = 3π/10 y = 4sin(5(3π/10)=4sin(3π/2) = -4
So in the first period, the max point is (π/10 , 4)
and the min point is (3π/10 , -4)
I used Calculus to find the max/min.
We could have done this by simply knowing something about our curve.
We know we have a sine curve of period 2π/5
So mark off a section of the line from 0 to 2π/5
and marking its 4 quadrants as
0 π/10 2π/10 3π/10 and 4π/10
or
0 , π/10 , π/5 , 3π/10 , and 2π/5
We know this sine curve starts at 0 goes up to 4 at π/10, back to 0 at π/5 , down to -4 at 3π/10 and then back up to 0 at 2π/5
you did not state in what domain you want these, so I just did the first period.
Of course each maximum happens again in each consecutive or previous period
so our max's happen at x = π/10 + k(2π/5) , where k is an integer
and our min's happen at x = 3π/10 + k(2π/5)
Of course each max will be 4 and each min will be -4
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