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Suppose r(t) = (et cos t)i + (et sin t)j. Show that the angle between r and a never changes. What is the angle?

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4 votes

Answer:

Angle = Ф =
cos^(-1)(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

Explanation:

Given vector r(t) =
e^(t)cost i + e^(t)sint j

As we know that,

velocity vector = v =
(dr)/(dt)

Implies that

velocity vector =
(e^(t) cost - e^(t) sint)i + (e^(t) sint - e^(t)cost )j

As acceleration is velocity over time so:

acceleration vector = a =
(dv)/(dt)

Implies that

vector a =


(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

vector a =
(-2e^(t)sint) i + ( 2e^(t)cost)j

Now scalar product of position vector r and acceleration vector a:

r. a =
<e^(t)cost, e^(t)sint> . <-2e^(t)sint, 2e^(t)cost>\\

r.a =
-2e^(2t)sintcost + 2e^(2t)sintcost

r.a = 0

Now, for angle between position vector r and acceleration vector a is given by:

cosФ =
(r.a)/(|r|.|a|) =
(0)/(|r|.|a|) = 0

Ф =
cos^(-1)(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

User Tokland
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