Prove that if a penguin walking in the plane has position vector always orthogonal to its velocity, then it is moving along a circular path. (Hint: consider the dot product of position with itself). Does your argument generalize to 3D?

a) r*dr/dt = 0 (dot product)

b) this argument can be generalized even to 3D

Step-by-step explanation:

for a circular path the relationship between x and y is

x²+y² = R² , where R= radius of the circle

for any point (x,y) the position vector is r and the velocity is the change in position with time , therefore v= dr/dt  

if we do the dot product of r and itself

r * r = (x,y)*(x,y) = x² + y² = R² ( constant)

r * r = R²

taking the derivative of r with respect to the time t

r * dr/dt + dr/dt * r = d(R²)/dt

2*r*dr/dt = 0

r*dr/dt = 0

therefore

r*dr/dt = 0

since the dot product of r and dr/dt is 0 they are orthogonal to each other.

For 3D , the equation of a sphere is

x²+y²+z³= R²

following the same steps than before we get r * r = (x,y,z)*(x,y,z) = x² + y² +z² = R²

and thus also r*dr/dt = 0

then this argument can be generalized even to 3D

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d

multiple all that

step-by-step explanation:

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