Velocity and Acceleration — Question 10

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Question 10

For a regular motion with speed v=ds/dtv=ds/dt, prove the Frenet acceleration formula a→=(dv/dt)𝑻+κv2𝑵\vec a=(dv/dt)\mathbf T+\kappa v^2\mathbf N. Identify the geometric role of each term.

Original worksheet page 1: question and worked solution for 6-11-010
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Question 10 – Solution

Strategy Write velocity as speed times unit tangent and apply the product and chain rules.

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Derivation v→=v𝑻.\vec v=v\mathbf T. Therefore a→=ddt(v𝑻)=dvdt𝑻+vd𝑻dsdsdt.\vec a=\frac{d}{dt}(v\mathbf T) =\frac{dv}{dt}\mathbf T+v\frac{d\mathbf T}{ds}\frac{ds}{dt}. Since d𝑻/ds=κ𝑵d\mathbf T/ds=\kappa\mathbf N and ds/dt=vds/dt=v, a→=dvdt𝑻+κv2𝑵.\boxed{\vec a=\frac{dv}{dt}\mathbf T+\kappa v^2\mathbf N}.

Meaning The tangential term changes speed. The normal term changes direction and points toward the local center of curvature; its magnitude grows quadratically with speed.

Original worksheet page 2: question and worked solution for 6-11-010

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