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Sagot :
Answer:
Angle = Ф = [tex]cos^{-1}[/tex](0) = 0
Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.
Step-by-step explanation:
Given vector r(t) = [tex]e^{t}cost i + e^{t}sint j[/tex]
As we know that,
velocity vector = v = [tex]\frac{dr}{dt}[/tex]
Implies that
velocity vector = [tex](e^{t} cost - e^{t} sint)i + (e^{t} sint - e^{t}cost )j[/tex]
As acceleration is velocity over time so:
acceleration vector = a = [tex]\frac{dv}{dt}[/tex]
Implies that
vector a =
[tex](e^{t}cost - e^{t}sint - e^{t}sint - e^{t}cost )i + ( e^{t}sint + e^{t}cost + e^{t}cost - e^{t}sint )j[/tex]
vector a = [tex](-2e^{t}sint) i + ( 2e^{t}cost)j[/tex]
Now scalar product of position vector r and acceleration vector a:
r. a = [tex]<e^{t}cost, e^{t}sint> . <-2e^{t}sint, 2e^{t}cost>\\[/tex]
r.a = [tex]-2e^{2t}sintcost + 2e^{2t}sintcost[/tex]
r.a = 0
Now, for angle between position vector r and acceleration vector a is given by:
cosФ = [tex]\frac{r.a}{|r|.|a|}[/tex] = [tex]\frac{0}{|r|.|a|} = 0[/tex]
Ф = [tex]cos^{-1}[/tex](0) = 0
Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.
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