Hey guys. in lecture 6 i got kinda confused on the proof that dr/ds is equal to the unit tangent vector. The professor stated that delta r is approximatley equal to t hat times delta s. Since t hat is 1, does that mean that dr=ds? could someone explain the proof to me?

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anonymous · Answer

The unit tangent vector is defined as the velocity vector divided by the length of the velocity vector (the speed):$$T ^{\rightarrow} = v ^{\rightarrow}/|v|$$Since $$|v ^{\rightarrow}|= ds/dt$$we can write $$T ^{\rightarrow}=v ^{\rightarrow}/(ds/dt) $$ or $$v ^{\rightarrow} = (ds/dt)T ^{\rightarrow}$$This makes sense: it merely says the velocity vector is a vector in the direction of the unit tangent vector, with length = speed. Now, the chain rule says that the velocity vector can be written as$$v ^{\rightarrow}=(dr ^{\rightarrow}/ds)(ds/dt)$$because cancellation of the common differential ds gives us back dr/dt. Thus we have $$(dr ^{\rightarrow}/ds)(ds/dt)=(ds/dt)T ^{\rightarrow}$$Elimination of (ds/dt) which appears on both sides gives  $$T ^{\rightarrow}=dr ^{\rightarrow}/ds$$ To answer your other question does $$dr ^{\rightarrow}=ds?$$ No!  We have $$\Delta r ^{\rightarrow} \approx T ^{\rightarrow} \Delta s$$ In the limit this relation becomes exact: $$dr ^{\rightarrow}=T ^{\rightarrow}(ds)$$Now don't confuse $$T ^{\rightarrow}$$ with $$|T ^{\rightarrow}|$$If the relation was $$dr ^{\rightarrow}=|T ^{\rightarrow}|ds$$ we would have dr=ds (because |T|=1 as you said) but we don't!  dr is a vector and ds is a scalar.  ds merely gives us the magnitude of the tiny vector displacement dr.  We still need information of which direction we are going.  This is given by the unit tangent vector T. So dr=T(dr)  tells us how far we are going =ds and which direction we are going = T.