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

idk

step-by-step explanation:

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true