Question

Prove that the normal component of acceleration is $0$at a point of inflection on the graph of a twice-differentiable function$y=f\left(x\right)$.

Short answer

Let Then,

role="math" localid="1649616080437"

We have . At a point of inflection we have$$\begin{gathered} a_T=\frac{||v\left(t\right)\times a\left(t\right)||}{||v\left(t\right)||} \\ {a}_T=0 \end{gathered}$$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

A twice-differentiable function is$y=f\left(x\right)$.

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

We know that the each inflection point $f\text{'}\text{'}\left(t\right)=0$.

Assume

Consider $v\left(t\right)\times a\left(t\right)$. Then,

Step 3. Determine the normal component of acceleration.

Consider the given question,

$$\begin{gathered} ||v\left(t\right)\times a\left(t\right)||=\sqrt{{\left(f\text{'}\text{'}\left(t\right)\right)}^2} \\ =f\text{'}\text{'}\left(t\right) \\ =0 \end{gathered}$$

At inflection point $f\text{'}\text{'}\left(t\right)=0$.

The normal component of acceleration is given below,

$$\begin{gathered} a_N=\frac{||v\left(t\right)\times a\left(t\right)||}{||v||} \\ =\frac{0}{||v||} \\ =0 \end{gathered}$$

Thus, the normal component of acceleration is $0$at a point of inflection of the graph of $f\left(x\right)$which is twice differentiable.