Answer:
The phase difference between yA and yB is 
Step-by-step explanation:
Given harmonic modeled as :
yA = 8 sin(2t - 
) And
yB = 8 sin(2t - 
)
The function as written as :
y = a sin(ωt - Ф) where Ф is phase difference
So , phase difference between yA and yB = ( Ф_1 - Ф_2 )
Or phase difference between yA and yB = ( - + )
Or, phase difference between yA and yB = 
I.e phase difference between yA and yB =
Hence The phase difference between yA and yB is Answer
<span>0, (4) = 4/9 <span>L circle = 2πR </span><span>L circle = 2 · π · 4/9 </span><span>L circle = 8π / 9 cm
</span></span>L = 2 * pi * R
<span>L = 2 * pi * 0. (4) cm = 0. (8) pi cm</span>
The answer
ellipse main equatin is as follow:
X²/ a² + Y²/ b² =1, where a≠0 and b≠0
for the first equation: <span>x = 3 cos t and y = 8 sin t
</span>we can write <span>x² = 3² cos² t and y² = 8² sin² t
and then </span>x² /3²= cos² t and y²/8² = sin² t
therefore, x² /3²+ y²/8² = cos² t + sin² t = 1
equivalent to x² /3²+ y²/8² = 1
for the second equation, <span>x = 3 cos 4t and y = 8 sin 4t we found
</span>x² /3²+ y²/8² = cos² 4t + sin² 4t=1
Answer: e^y=x
ln(x)=y
e^ln(x)=e^y
x=e^y
