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Prove that the normal component of acceleration is 0at a point of inflection on the graph of a twice-differentiable functiony=f(x).

Short Answer

Expert verified

Let rt=t,ft,0.Then,

role="math" localid="1649616080437" vt=1,f't,0,at=0,f''t,0

We have vt×at=0,0,f''t. At a point of inflection we haveaT=vt×atvtaT=0.

Step by step solution

01

Step 1. Given information.

Consider the given question,

A twice-differentiable function isy=f(x).

02

Step 2. Consider each inflection point f''t=0.

We know that the each inflection point f''t=0.

Assume rt=t,ft,0vt=r'tdtvt=t,f't,0at=v'tdtat=0,f''t,0

Consider vt×at. Then,

vt×at=ijk1f't00f''t0=i0-j0+kf''t=0,0,f''t

03

Step 3. Determine the normal component of acceleration.

Consider the given question,

vt×at=f''t2=f''t=0

At inflection point f''t=0.

The normal component of acceleration is given below,

aN=vt×atv=0v=0

Thus, the normal component of acceleration is 0at a point of inflection of the graph of fxwhich is twice differentiable.

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