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Chapter 10: Problem 38

Find the curvature and radius of curvature of the plane curve at the given value of \(x\). $$ y=m x+b, \quad x=a $$

Short Answer

Expert verified
The curvature of the given curve at any point is 0 and the radius of curvature is infinity.

Step by step solution

01

Find the derivative of the curve

The given equation of the curve is \(y = mx + b\). We first need to find its derivative which is represented as \(y'\). Using differentiation, we find that \(y' = m\) which means the slope of the line is a constant that doesn't change with respect to \(x\).
02

Calculate the 2nd derivative of the curve

The second derivative of the curve, represented as \(y''\), gives us the rate of change of the slope. In our case, since the first derivative is a constant, the 2nd derivative for our function is zero, \(y'' = 0\).
03

Apply the formula for curvature

The formula for the curvature, represented as \(k\), is given by \(k = |y''|/(1+{(y')}^2)^{\frac{3}{2}}\). Substituting the values of the first and second derivatives from steps 1 and 2, we have \(k = |0|/(1+{(m)}^2)^{\frac{3}{2}} = 0\). Thus, the curvature is 0.
04

Calculate the radius of curvature

The radius of curvature, represented as \(R\), is the reciprocal of curvature, \(R = 1/k\). However, here, the curvature is 0. Therefore, it's not possible to calculate the radius of curvature as it would result in division by zero, and we can say that the radius of curvature is infinity.

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Most popular questions from this chapter

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. $$ D_{t}\left[\mathbf{r}(t) \times \mathbf{r}^{\prime}(t)\right]=\mathbf{r}(t) \times \mathbf{r}^{\prime \prime}(t) $$

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