If a curve has the property that the position vector

Julia White 2021-12-20 Answered
If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r(t), show that the curve lies on a sphere with center the origin.
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Expert Answer

kaluitagf
Answered 2021-12-21 Author has 38 answers

Step 1
It is given that r(t)×r(t)=0 for all tR
Now, let us consider,
d(r(t)×r(t))dt=r(t)×r(t)+r(t)×r(t)
=2×r(t)×r(t)
(given r(t)×r(t)=0)
=0
This implies that d(r(t)2)dt=0
⇒∥r(t)2 is constant and therefore, r(t) should also be a constant. Let K be the constant such that r(t)∥=K
This implies that, r(t)2=K2
Let us say x, y, z are the components of vector function r(t), this implies that, we have,
x2+y2+z2=K2
which is a sphere centered at (0, 0, 0) and with radius K
Thus, we have r(t)=x(t), y(t), z(t) such that,
r(t)∥=x2+y2+z2
⇒∥r(t)2=x2+y2+z2
Hence, the curve r(t) must be a sphere centered at origin.

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zurilomk4
Answered 2021-12-22 Author has 35 answers
Step 1
Given that if a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r(t). It is required to show that the curve lies on a sphere with centre the origin.
Let r(t) be r(t)=x(t), y(t), z(t) then,
|r(t)|=x2+y2+z2
|r(t)|2=x2+y2+z2
Step 2
Since the position vector r(t) is always perpendicular to the tangent vector r(t)
r(t)×r(t)=0
2×r(t)×r(t)=2×0
2×r(t)×r(t)=0
ddt(|r(t)|2)=0
Now integrate on both sides,
ddt(|r(t)|2)=0
|r(t)|2=c
It is observed that as |r(t)|=c then x2+y2+z2=c, the position vector of the curve is a constant that is the distance is same in all directions.
Thus, the curve lies on a sphere whose centre is origin.

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nick1337
Answered 2021-12-27 Author has 575 answers

Step 1
So we have
r(t)×r(t)=0
which gives us;
(r(t)×r(t))=2r(t)×r(t)=0
Thus:
r(t)2=r(t)×r(t)=c
for some constant c
So the curve lies on:
x2+y2+z2=c

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