Analyzing Bee's Spiral Path: Velocity & Acceleration

[chaotixmonjuish]

1.)

A bee goes from its hive in a spiral path given in plane polar coordinates by
r = b*e^kt , θ = ct,
where b, k, c are positive constants. Show that the angle between the velocity vector and the
acceleration vector remains constant as the bee moves outward. (Hint: Find v · a/va.)

so here is my v and a

2.)
v = (r')e_r+(r*θ')e_θ

a = (r''+rθ')e_r+(rθ''+2r'θ')e_θ

r' = bk*e^kt
r'' = bk²e^kt
θ' = c

3.) my attempt at a solution

bke^kt(bk²e^kt-be^ktc²)+(be^ktc)(2bke^ktc)

is that the right dot product

this is where I'm stuck

 

[chaotixmonjuish]

so I got down to this:

e^2kt(a bunch of constants)+e^2kt(a bunch of constants)/e^4kt


all the e's canceled out and left just constants

 

[markpace123]

My answer is for the dot product is:

e^2kt(b²k³ + b²kc + 2b²kc²)

 

[markpace123]

you continue by finding the modulus of v and a then by using the hint you cross out the

b²e^2kt and end up with cos^-1(a bunch of constants) and the answer

 

[asteriatic]

.

I would first clarify the question and make sure I understand what is being asked. From the given information, it seems like we are trying to prove that the angle between the velocity vector and the acceleration vector remains constant as the bee moves outward in its spiral path.

To do this, we can use the dot product formula:

v · a = |v| |a| cosθ

where θ is the angle between the two vectors.

From the given equations, we can calculate the magnitude of the velocity vector as:

|v| = √(v · v) = √[(bk*ekt)^2 + (bektc)^2] = bk*ekt

Similarly, the magnitude of the acceleration vector is:

|a| = √(a · a) = √[(bk2ekt + bektc^2)^2 + (2bkektc)^2] = √(bk2ekt + bektc^2)^2 + 4b2k2e2ktc^2] = bk2ekt + 2bkektc

Now, to find the angle θ between the two vectors, we can use the dot product formula again:

cosθ = (v · a)/(|v| |a|)

Substituting the values we calculated above:

cosθ = [(bkekt)(bk2ekt + 2bkektc)]/(bk*ekt)(bk2ekt + 2bkektc) = 1

This shows that the angle between the velocity vector and the acceleration vector is always cos^-1(1) = 0, which means it remains constant as the bee moves outward in its spiral path.

Therefore, we have proven that the angle between the velocity vector and the acceleration vector remains constant.