MHB Vectors and their derivative proof

[brunette15]

The question suggests that r(t) = (x(t),y(t),z(t)) is a position vector along some curve where t goes from negative to positive infinity. Now suppose t has been chosen so that 1 = the dot product of dr/dt and dr/dt. Show that 0 = the dot product of dr/dt and d^2r/dt^2.

I have attempted to expand the dot products then regroup them but I am having a bit of trouble with this :(

 

[I like Serena]

The question suggests that r(t) = (x(t),y(t),z(t)) is a position vector along some curve where t goes from negative to positive infinity. Now suppose t has been chosen so that 1 = the dot product of dr/dt and dr/dt. Show that 0 = the dot product of dr/dt and d^2r/dt^2.

I have attempted to expand the dot products then regroup them but I am having a bit of trouble with this :(

Hi brunette15! Welcome to MHB! (Smile)

Let's denote $\dot r = \d r t$ and $\ddot r = \frac{d^2r}{dt^2}$.
This is a common notation to denote time derivatives.

Then we have:
$$\dot r \cdot \dot r = 1\tag 1$$
and we want to prove that:
$$\dot r \cdot \ddot r \overset{?}{=} 0$$

What do you get if you take the derivative of $(1)$?
That is:
$$\frac{d}{dt}(\dot r \cdot \dot r)$$
(Wondering)

 

[brunette15]

Hi brunette15! Welcome to MHB! (Smile)

Let's denote $\dot r = \d r t$ and $\ddot r = \frac{d^2r}{dt^2}$.
This is a common notation to denote time derivatives.

Then we have:
$$\dot r \cdot \dot r = 1\tag 1$$
and we want to prove that:
$$\dot r \cdot \ddot r \overset{?}{=} 0$$

What do you get if you take the derivative of $(1)$?
That is:
$$\frac{d}{dt}(\dot r \cdot \dot r)$$
(Wondering)

Thanks IlikeSerena! I am still struggling to see a connection. Would $$\frac{d}{dt}(\dot r \cdot \dot r)$$ just equal 0 then?

 

[brunette15]

Hi brunette15! Welcome to MHB! (Smile)

Let's denote $\dot r = \d r t$ and $\ddot r = \frac{d^2r}{dt^2}$.
This is a common notation to denote time derivatives.

Then we have:
$$\dot r \cdot \dot r = 1\tag 1$$
and we want to prove that:
$$\dot r \cdot \ddot r \overset{?}{=} 0$$

What do you get if you take the derivative of $(1)$?
That is:
$$\frac{d}{dt}(\dot r \cdot \dot r)$$
(Wondering)

Thankyou! I am still struggling to see a connection however... :/

 

[I like Serena]

Thanks IlikeSerena! I am still struggling to see a connection. Would $$\frac{d}{dt}(\dot r \cdot \dot r)$$ just equal 0 then?

Yes. (Nod)

Thankyou! I am still struggling to see a connection however... :/

The product rule for derivatives says:
$$\d {} x (f(x) \cdot g(x)) = f'(x)g(x) + f(x)g'(x)$$

Similarly we can expect that:
$$\d {} t (\mathbf{\dot r} \cdot \mathbf{\dot r}) = \mathbf{\ddot r} \cdot \mathbf{\dot r} + \mathbf{\dot r} \cdot \mathbf{\ddot r} = 2 \mathbf{\dot r} \cdot \mathbf{\ddot r} = 0$$
which is what we need to prove.

That leaves showing that this is actually the case. (Thinking)

 

[brunette15]

Yes. (Nod)
The product rule for derivatives says:
$$\d {} x (f(x) \cdot g(x)) = f'(x)g(x) + f(x)g'(x)$$

Similarly we can expect that:
$$\d {} t (\mathbf{\dot r} \cdot \mathbf{\dot r}) = \mathbf{\ddot r} \cdot \mathbf{\dot r} + \mathbf{\dot r} \cdot \mathbf{\ddot r} = 2 \mathbf{\dot r} \cdot \mathbf{\ddot r} = 0$$
which is what we need to prove.

That leaves showing that this is actually the case. (Thinking)

Thankyou so much!