Here’s a problem (2.1.5, p. 23) from Lang’s “A Second Course in Calculus”:

In exercises 3 and 4, show that the velocity vector is perpendicular to the position vector.

For reference, exercise 3 is: (cos(*t*), sin(*t*)), and exercise 4 is (cos(3*t*), sin(3*t*)). If a velocity vector is perpendicular to a position vector, then the dot product of the position and velocity vectors has to be 0.

Let’s start with the position vector in exercise 3: (cos(*t*), sin(*t*)). The velocity vector of this position vector is the derivative of each term, or (-sin(*t*), cos(*t*)). Let’s take the dot product:

(cos(*t*))(-sin(*t*)) + (sin(*t*))(cos(*t*)) = -(sin(*t*)cos(*t*)) + (sin(*t*)cos(*t*) = 0

Thus, in exercise 3, the existence of a dot product of 0 confirms that the velocity vector is perpendicular to the position vector.

Now, for exercise 5, the position vector is (cos(3*t*), sin(3*t*)), and, therefore, the velocity vector is:

Therefore, the velocity vector of (cos(3*t*), sin(3*t*)) is (-3sin(3*t*), 3cos(3*t*)). Taking the dot product, we have:

(cos(3*t*))(-3sin(3*t*)) + (sin(3*t*)(3cos(3*t*)) =

-3(cos(3*t*))(sin(3*t*)) + 3(cos(3*t*)(sin(3*t*)) = 0

Thus, in exercise 5 also, the existence of a dot product of 0 confirms that the velocity vector is perpendicular to the position vector.

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