Question

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?

Answer 1

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.

Step-by-step explanation:

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

As we know that,

velocity vector = v = $\frac{dr}{dt}$

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 = $\frac{dv}{dt}$

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 = $.$

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 = $\frac{r.a}{|r|.|a|}$ = $\frac{0}{|r|.|a|} = 0$

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.