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Chapter 11: Problem 3

Express the arc length of a curve in terms of the speed of an object moving along the curve.

Short Answer

Expert verified
Question: Express the speed of an object moving along a curve in terms of the arc length of the curve. Answer: The speed (v) of an object moving along a curve can be expressed in terms of the arc length as follows: v = 鈭�((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)

Step by step solution

01

Write the arc length formula

For a curve represented as a function of a parameter t, the arc length (s) can be found using the following formula: s = Integral[||d**r|| dt] Here, r(t) is a position vector of a point on the curve, ||d**r|| is the magnitude of the curve's tangent vector, and t is the parameter.
02

Find the tangent vector

The tangent vector at a point on the curve is given by the derivative of the position vector of that point with respect to the parameter t. Tangent vector = d**r/dt
03

Find the magnitude of the tangent vector

To find the magnitude (length) of the tangent vector, take the square root of the sum of the squares of its components, i.e., ||d**r|| = 鈭�((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) Here, x(t), y(t), z(t) are the components of the position vector r(t).
04

Write arc length in terms of magnitude of the tangent vector

Now, substitute the expression for the magnitude of the tangent vector into the arc length formula: s = Integral[鈭�((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)] dt
05

Differentiate the arc length with respect to time

To express the arc length in terms of the speed, differentiate the arc length (s) with respect to time (t): ds/dt = d[鈭�((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)] dt The derivative of the arc length with respect to time (ds/dt) is the speed (v) of the object moving along the curve. Therefore, we have: Speed (v) = 鈭�((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) This gives us the expression for the speed of an object moving along the curve in terms of the arc length.

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Most popular questions from this chapter

Prove the following vector properties using components. Then make a sketch to illustrate the property geometrically. Suppose \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in the \(x y\) -plane and a and \(c\) are scalars. $$a(c \mathbf{v})=(a c) \mathbf{v}$$

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=2^{t} \mathbf{i}+\frac{1}{1+2 t} \mathbf{j}+\ln t \mathbf{k}$$

A pair of nonzero vectors in the plane is linearly dependent if one vector is a scalar multiple of the other. Otherwise, the pair is linearly independent. a. Which pairs of the following vectors are linearly dependent and which are linearly independent: \(\mathbf{u}=\langle 2,-3\rangle\) \(\mathbf{v}=\langle-12,18\rangle,\) and \(\mathbf{w}=\langle 4,6\rangle ?\) b. Geometrically, what does it mean for a pair of nonzero vectors in the plane to be linearly dependent? Linearly independent? c. Prove that if a pair of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is linearly independent, then given any vector \(w\), there are constants \(c_{1}\) and \(c_{2}\) such that \(\mathbf{w}=c_{1} \mathbf{u}+c_{2} \mathbf{v}\)

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Triangle Inequality Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\) b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\) c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Algebra inequality Show that $$\left(u_{1}+u_{2}+u_{3}\right)^{2} \leq 3\left(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\right)$$ for any real numbers \(u_{1}, u_{2},\) and \(u_{3} .\) (Hint: Use the CauchySchwarz Inequality in three dimensions with \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and choose v in the right way.)
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