Is dr/dt Always Perpendicular to r for a Unit Vector r(t)?

[Liquidxlax]

Homework Statement

4. Let r=r(t) be a vector whose length is always 1 (it may vary in direction). Prove that either r is a constant vector or dr/dt is perpendicular to r. hit: differentiate r.r

Homework Equations

basic derivatives with respect to t for velocity and acceleration

The Attempt at a Solution

r = 1 = x + y + z

df = d(r.r) = 2x(df/dx) + 2y(df/dy) + 2z(df/dz)

don't have notes on it yets, so I'm not sure where to go from here.

 

[fluxions]

One can write \vec{r} = r \hat{r}. You are given that the length is always 1, so what can you conclude about r?

Then note that \vec{r} \cdot \vec{r} = r \hat{r} \cdot r \hat{r} = r^2. Insert the value of r you figured out above, and differentiate this expression with respect to time. You should end up with "thing1 dot thing2 = thing3". What can you conclude about either thing1 or thing2?

 

[HallsofIvy]

I presume that by "r= x+ y+ z" you mean
\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}

but then that cannot be equal to the number 1!

In any case, you don't want to reduce to components.

Product rule:
\frac{d \vec{r}\cdot\vec{r}}{dt}= \frac{d\vec{r}}{dt}\cdot\vec{r}+ \vec{r}\cdot\frac{d\vec{r}}{dt}= \frac{d(1)}{dt}= 0

 

[Liquidxlax]

I presume that by "r= x+ y+ z" you mean
\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}

but then that cannot be equal to the number 1!

In any case, you don't want to reduce to components.

Product rule:
\frac{d \vec{r}\cdot\vec{r}}{dt}= \frac{d\vec{r}}{dt}\cdot\vec{r}+ \vec{r}\cdot\frac{d\vec{r}}{dt}= \frac{d(1)}{dt}= 0

okay, thanks that seems to make sense