Calculus · Chapter 13

Vector Functions

Up to now the functions we worked with returned numbers. This chapter introduces functions that return vectors. They are the natural language for describing curves and surfaces in space and for tracking how objects move through it — culminating in a calculus-based derivation of Kepler's laws of planetary motion.

Reference adaptation based on the source course, Chapter 13. Concepts are paraphrased and condensed; all figures are original schematic redraws. Exercises and review problems are omitted.

Section 13.1

Vector Functions and Space Curves

A vector function takes a real number and returns a vector. The number is usually time, which is why these functions are the right tool for describing motion and curves in three dimensions.

Vector function

A function whose domain is a set of real numbers and whose range is a set of vectors. In space, a vector function returns a three-dimensional vector for each input \(t\) in its domain.

Component functions

The three real-valued functions \(f\), \(g\), \(h\) that supply the \(x\), \(y\), \(z\) coordinates of a vector function. Everything about a vector function reduces to its components, so limits, derivatives, and integrals are all computed component by component.

Vector function in components

\[ \mathbf{r}(t)=\langle f(t),\,g(t),\,h(t)\rangle = f(t)\,\mathbf{i}+g(t)\,\mathbf{j}+h(t)\,\mathbf{k} \]

The domain of \(\mathbf{r}\) is the set of \(t\) for which every component is defined — that is, the intersection of the three component domains.

EXAMPLE 1 Finding the domain

Find the domain of \(\mathbf{r}(t)=\langle t^{3},\,\ln(3-t),\,\sqrt{t}\,\rangle\).

Solution

\(t^{3}\) is defined everywhere; \(\ln(3-t)\) requires \(3-t>0\), i.e. \(t<3\); \(\sqrt{t}\) requires \(t\ge 0\).

Intersecting the three conditions gives the domain \(\;[0,\,3)\).

Limits and continuity

Because a vector function is just three real functions bundled together, its limit is taken in each coordinate separately.

Limit of a vector function

\[ \lim_{t\to a}\mathbf{r}(t)=\left\langle \lim_{t\to a}f(t),\;\lim_{t\to a}g(t),\;\lim_{t\to a}h(t)\right\rangle \]

Continuity

A vector function \(\mathbf{r}\) is continuous at \(a\) when \(\lim_{t\to a}\mathbf{r}(t)=\mathbf{r}(a)\). This holds exactly when all three component functions are continuous at \(a\).

EXAMPLE 2 A component-wise limit

Evaluate \(\displaystyle\lim_{t\to 0}\mathbf{r}(t)\) for \(\mathbf{r}(t)=\left\langle (1+t^{3}),\;t e^{t},\;\dfrac{\sin t}{t}\right\rangle\).

Solution

Take the limit in each coordinate: \(\lim_{t\to0}(1+t^{3})=1\), \(\;\lim_{t\to0}t e^{t}=0\).

The standard limit \(\lim_{t\to0}\frac{\sin t}{t}=1\) gives the third component.

Therefore \(\displaystyle\lim_{t\to 0}\mathbf{r}(t)=\langle 1,\,0,\,1\rangle=\mathbf{i}+\mathbf{k}\).

Space curves

Space curve

The set of points \((f(t),g(t),h(t))\) traced as \(t\) runs over an interval, where the components are continuous. The equations \(x=f(t),\,y=g(t),\,z=h(t)\) are its parametric equations and \(t\) is the parameter.

Key idea

Picture a particle whose position at time \(t\) is \(\mathbf{r}(t)\). The curve is the trail swept out by the tip of that moving position vector. A curve and "a moving vector that draws it" are two views of the same object.

Figure 13.1.1The curve \(C\) is traced out by the tip of the moving position vector \(\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle\).

EXAMPLE 3 A line as a vector function

Describe the curve \(\mathbf{r}(t)=\langle 1+t,\;2+5t,\;-1+6t\rangle\).

Solution

The parametric equations \(x=1+t,\;y=2+5t,\;z=-1+6t\) are those of a line.

It passes through \((1,2,-1)\) and is parallel to \(\langle 1,5,6\rangle\); equivalently \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\) with \(\mathbf{r}_0=\langle1,2,-1\rangle\), \(\mathbf{v}=\langle1,5,6\rangle\).

EXAMPLE 4 The circular helix

Sketch the curve \(\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+t\,\mathbf{k}\).

Solution

Since \(x^{2}+y^{2}=\cos^{2}t+\sin^{2}t=1\), the curve lies on the cylinder \(x^{2}+y^{2}=1\).

The point \((x,y,0)\) circles counterclockwise around the unit circle while \(z=t\) rises steadily.

The result spirals upward around the cylinder — a helix.

Figure 13.1.2A helix winds around the cylinder \(x^{2}+y^{2}=1\) while its height grows linearly with \(t\). The same shape appears in coiled springs and in the double-helix model of DNA.

Admin Note

The helix is worth memorizing as a "base case." Whenever a problem mixes circular motion in two coordinates with steady growth in the third, expect a helix. In real hardware this is exactly a spring or a screw thread — the pitch of the thread is the \(z\) gained per full turn.

EXAMPLE 5 Parametrizing a line segment

Find a vector equation for the segment from \(P(1,3,-2)\) to \(Q(2,-1,3)\).

Solution

Use the segment formula \(\mathbf{r}(t)=(1-t)\,\mathbf{r}_0+t\,\mathbf{r}_1,\;0\le t\le1\) with \(\mathbf{r}_0=\langle1,3,-2\rangle\), \(\mathbf{r}_1=\langle2,-1,3\rangle\).

This gives \(\mathbf{r}(t)=\langle 1+t,\;3-4t,\;-2+5t\rangle,\;0\le t\le1\).

Parametric form: \(x=1+t,\;y=3-4t,\;z=-2+5t\).

EXAMPLE 6 Curve of intersection

Find a vector function for the curve where the cylinder \(x^{2}+y^{2}=1\) meets the plane \(y+z=2\).

Solution

The projection onto the \(xy\)-plane is the unit circle, so \(x=\cos t,\;y=\sin t\).

The plane gives \(z=2-y=2-\sin t\).

Hence \(\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+(2-\sin t)\,\mathbf{k},\;0\le t\le 2\pi\) — an ellipse traced on the cylinder.

Strategy

To parametrize a curve of intersection, parametrize the "easy" projection first (often a circle from a cylinder), then substitute into the second surface's equation to recover the remaining coordinate.

Famous space curves illustrate why technology helps: the toroidal spiral lies on a torus, the trefoil knot crosses over itself, and the twisted cubic \(\mathbf{r}(t)=\langle t,t^{2},t^{3}\rangle\) is clearest when viewed inside a bounding box rather than against bare axes.

Section 13.2

Derivatives and Integrals of Vector Functions

The calculus of vector functions mirrors single-variable calculus exactly, one component at a time. The derivative carries an extra geometric payload: it is the tangent vector to the curve.

Derivative of a vector function

Defined by the same difference-quotient limit as for real functions. Geometrically, the secant vector \(\mathbf{r}(t+h)-\mathbf{r}(t)\), scaled by \(1/h\), approaches a vector along the tangent line as \(h\to0\).

Definition of the derivative

\[ \mathbf{r}'(t)=\frac{d\mathbf{r}}{dt}=\lim_{h\to 0}\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} \]

Tangent vector and unit tangent

\(\mathbf{r}'(t)\) is the tangent vector at the point (provided it exists and is nonzero). Dividing by its length gives the unit tangent vector, which records direction of travel without magnitude.

Unit tangent vector

\[ \mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\lvert \mathbf{r}'(t)\rvert} \]

Componentwise differentiation

If \(\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle\) with differentiable components, then \(\mathbf{r}\) is differentiable and its derivative is obtained by differentiating each component:

\[ \mathbf{r}'(t)=\langle f'(t),\,g'(t),\,h'(t)\rangle \]

Figure 13.2.1As \(Q\to P\), the secant vector swings into the tangent direction \(\mathbf{r}'(t)\), which lies along the tangent line at \(P\).

EXAMPLE 1 Derivative and unit tangent

For \(\mathbf{r}(t)=(1+t^{3})\,\mathbf{i}+t e^{-t}\,\mathbf{j}+\sin 2t\,\mathbf{k}\), find \(\mathbf{r}'(t)\) and the unit tangent at \(t=0\).

Solution

Differentiate each component: \(\mathbf{r}'(t)=3t^{2}\,\mathbf{i}+(1-t)e^{-t}\,\mathbf{j}+2\cos 2t\,\mathbf{k}\).

At \(t=0\): \(\mathbf{r}'(0)=\mathbf{j}+2\mathbf{k}\), with length \(\sqrt{1+4}=\sqrt5\).

So \(\mathbf{T}(0)=\dfrac{1}{\sqrt5}\,\mathbf{j}+\dfrac{2}{\sqrt5}\,\mathbf{k}\).

EXAMPLE 3 Tangent line to a helix

Find parametric equations for the tangent line to \(x=2\cos t,\;y=\sin t,\;z=t\) at \(\left(0,1,\tfrac{\pi}{2}\right)\).

Solution

\(\mathbf{r}'(t)=\langle -2\sin t,\;\cos t,\;1\rangle\).

The point corresponds to \(t=\tfrac{\pi}{2}\), giving tangent vector \(\langle -2,0,1\rangle\).

Line through \(\left(0,1,\tfrac{\pi}{2}\right)\): \(x=-2t,\;y=1,\;z=\tfrac{\pi}{2}+t\).

Differentiation rules

Every product rule from single-variable calculus has a vector counterpart — including two distinct "product" rules, one for the dot product and one for the cross product (where order must be preserved).

Differentiation rules for vector functions

For differentiable \(\mathbf{u},\mathbf{v}\), scalar \(c\), and scalar function \(f\):

\[ \begin{aligned} &\tfrac{d}{dt}[\mathbf{u}+\mathbf{v}]=\mathbf{u}'+\mathbf{v}' \qquad \tfrac{d}{dt}[c\,\mathbf{u}]=c\,\mathbf{u}'\\[4pt] &\tfrac{d}{dt}[f\mathbf{u}]=f'\mathbf{u}+f\mathbf{u}'\\[4pt] &\tfrac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\mathbf{u}'\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{v}'\\[4pt] &\tfrac{d}{dt}[\mathbf{u}\times\mathbf{v}]=\mathbf{u}'\times\mathbf{v}+\mathbf{u}\times\mathbf{v}'\\[4pt] &\tfrac{d}{dt}\big[\mathbf{u}(f(t))\big]=f'(t)\,\mathbf{u}'(f(t)) \quad(\text{chain rule}) \end{aligned} \]

EXAMPLE 4 Constant length forces orthogonality

Show that if \(\lvert\mathbf{r}(t)\rvert=c\) is constant, then \(\mathbf{r}'(t)\) is orthogonal to \(\mathbf{r}(t)\) for all \(t\).

Solution

Since \(\mathbf{r}\cdot\mathbf{r}=\lvert\mathbf{r}\rvert^{2}=c^{2}\) is constant, differentiate both sides.

The dot-product rule gives \(0=\mathbf{r}'\cdot\mathbf{r}+\mathbf{r}\cdot\mathbf{r}'=2\,\mathbf{r}'\cdot\mathbf{r}\).

Thus \(\mathbf{r}'\cdot\mathbf{r}=0\): the tangent is perpendicular to the radius. Geometrically, a curve on a sphere centered at the origin always has its tangent perpendicular to the position vector.

Admin Note

Example 4 is a workhorse identity in dynamics. "Speed constant ⇒ acceleration perpendicular to velocity" is the same statement applied to \(\mathbf{v}\) instead of \(\mathbf{r}\). It is why uniform circular motion has purely centripetal acceleration — keep this in your back pocket for Section 13.4.

Integrals

Integration is again componentwise, and the Fundamental Theorem of Calculus extends directly.

Definite integral, componentwise

\[ \int_a^b \mathbf{r}(t)\,dt=\left(\int_a^b f\,dt\right)\mathbf{i}+\left(\int_a^b g\,dt\right)\mathbf{j}+\left(\int_a^b h\,dt\right)\mathbf{k} \]

Fundamental Theorem for vector functions

\[ \int_a^b \mathbf{r}(t)\,dt=\mathbf{R}(b)-\mathbf{R}(a),\qquad \mathbf{R}'(t)=\mathbf{r}(t) \]

EXAMPLE 5 Integrating component by component

For \(\mathbf{r}(t)=2\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+2t\,\mathbf{k}\), find the indefinite integral and \(\int_0^{\pi/2}\mathbf{r}(t)\,dt\).

Solution

Indefinite: \(\int\mathbf{r}\,dt = 2\sin t\,\mathbf{i}-\cos t\,\mathbf{j}+t^{2}\,\mathbf{k}+\mathbf{C}\).

Evaluate from \(0\) to \(\tfrac{\pi}{2}\): \(\big[2\sin t\,\mathbf{i}-\cos t\,\mathbf{j}+t^{2}\,\mathbf{k}\big]_0^{\pi/2}\).

Result: \(2\,\mathbf{i}+\mathbf{j}+\dfrac{\pi^{2}}{4}\,\mathbf{k}\).

Section 13.3

Arc Length and Curvature

Two geometric measurements dominate this section: how long a curve is, and how sharply it bends. Both are built from the tangent vector and are designed to be independent of how the curve is parametrized.

Arc length

The length of a space curve is the limit of lengths of inscribed polygons. For a smoothly traced curve it equals the integral of the speed \(\lvert\mathbf{r}'(t)\rvert\) over the parameter interval.

Arc length of a space curve

\[ L=\int_a^b \lvert \mathbf{r}'(t)\rvert\,dt=\int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^{2}+\left(\tfrac{dy}{dt}\right)^{2}+\left(\tfrac{dz}{dt}\right)^{2}}\;dt \]

Why it's parametrization-independent

Different vector functions can trace the same curve, yet the arc-length integral always returns the same number. Length belongs to the curve, not to the clock used to walk it.

EXAMPLE 1 Length of one turn of a helix

Find the length of \(\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+t\,\mathbf{k}\) from \((1,0,0)\) to \((1,0,2\pi)\).

Solution

\(\mathbf{r}'(t)=\langle -\sin t,\cos t,1\rangle\), so \(\lvert\mathbf{r}'(t)\rvert=\sqrt{\sin^2t+\cos^2t+1}=\sqrt2\).

The arc covers \(0\le t\le 2\pi\).

\(L=\displaystyle\int_0^{2\pi}\sqrt2\,dt=2\sqrt2\,\pi\).

The arc length function and reparametrization

Arc length function

\(s(t)=\int_a^t \lvert\mathbf{r}'(u)\rvert\,du\) gives the distance traveled along the curve up to parameter \(t\). By the Fundamental Theorem, \(\dfrac{ds}{dt}=\lvert\mathbf{r}'(t)\rvert\) — the speed.

Parametrization by arc length

Solving \(s=s(t)\) for \(t\) and substituting expresses the curve in terms of distance traveled. This natural parameter depends only on the curve's shape, not on any coordinate system — which is exactly why curvature is defined with respect to it.

Curvature

A measure of how quickly the curve changes direction. It is the magnitude of the rate of change of the unit tangent vector with respect to arc length. A straight line has curvature \(0\); a tightly turning curve has large curvature.

Curvature (definition)

\[ \kappa=\left\lvert \frac{d\mathbf{T}}{ds}\right\rvert \]

Curvature in terms of the parameter t

\[ \kappa(t)=\frac{\lvert \mathbf{T}'(t)\rvert}{\lvert \mathbf{r}'(t)\rvert} \]

Cross-product formula for curvature

For a curve given by \(\mathbf{r}(t)\), the curvature can be computed without finding \(\mathbf{T}\):

\[ \kappa(t)=\frac{\lvert \mathbf{r}'(t)\times \mathbf{r}''(t)\rvert}{\lvert \mathbf{r}'(t)\rvert^{3}} \]

EXAMPLE 3 Curvature of a circle

Show that a circle of radius \(a\) has curvature \(1/a\).

Solution

Parametrize: \(\mathbf{r}(t)=a\cos t\,\mathbf{i}+a\sin t\,\mathbf{j}\), so \(\mathbf{r}'(t)=\langle-a\sin t,a\cos t\rangle\) and \(\lvert\mathbf{r}'\rvert=a\).

\(\mathbf{T}(t)=\langle -\sin t,\cos t\rangle\), so \(\mathbf{T}'(t)=\langle-\cos t,-\sin t\rangle\) with \(\lvert\mathbf{T}'\rvert=1\).

\(\kappa=\dfrac{\lvert\mathbf{T}'\rvert}{\lvert\mathbf{r}'\rvert}=\dfrac{1}{a}\). Small circles bend sharply; large circles are nearly straight.

Curvature of a plane curve y = f(x)

\[ \kappa(x)=\frac{\lvert f''(x)\rvert}{\big[\,1+(f'(x))^{2}\,\big]^{3/2}} \]

EXAMPLE 4 Curvature of the twisted cubic

Find the curvature of \(\mathbf{r}(t)=\langle t,t^{2},t^{3}\rangle\) at a general point and at the origin.

Solution

\(\mathbf{r}'(t)=\langle1,2t,3t^{2}\rangle\), \(\;\mathbf{r}''(t)=\langle0,2,6t\rangle\).

\(\mathbf{r}'\times\mathbf{r}''=\langle 6t^{2},-6t,2\rangle\), so \(\lvert\mathbf{r}'\times\mathbf{r}''\rvert=2\sqrt{9t^{4}+9t^{2}+1}\) and \(\lvert\mathbf{r}'\rvert=\sqrt{1+4t^{2}+9t^{4}}\).

\(\kappa(t)=\dfrac{2\sqrt{9t^{4}+9t^{2}+1}}{(1+4t^{2}+9t^{4})^{3/2}}\). At the origin \((t=0)\), \(\kappa=2\).

Normal and binormal vectors

Principal unit normal vector N

Since \(\mathbf{T}\) has constant length \(1\), its derivative \(\mathbf{T}'\) is orthogonal to \(\mathbf{T}\). Normalizing \(\mathbf{T}'\) gives \(\mathbf{N}=\mathbf{T}'/\lvert\mathbf{T}'\rvert\), pointing in the direction the curve is turning.

Binormal vector B and the TNB frame

\(\mathbf{B}=\mathbf{T}\times\mathbf{N}\) is a unit vector perpendicular to both. Together \(\mathbf{T},\mathbf{N},\mathbf{B}\) form a moving orthonormal frame that travels along the curve — central to differential geometry and to spacecraft attitude control.

Figure 13.3.1The three mutually orthogonal unit vectors of the moving frame: tangent \(\mathbf{T}\) (direction of travel), normal \(\mathbf{N}\) (direction of turning), and binormal \(\mathbf{B}=\mathbf{T}\times\mathbf{N}\).

Normal, osculating planes and the osculating circle

The normal plane at a point is spanned by \(\mathbf{N}\) and \(\mathbf{B}\); the osculating plane is spanned by \(\mathbf{T}\) and \(\mathbf{N}\) and is the plane that hugs the curve most closely there. The osculating circle lies in that plane, shares the curve's tangent and curvature, and has radius \(1/\kappa\) — the best-fitting circle to the curve at that point.

EXAMPLE 6 & 7 Frame and planes for a helix

For \(\mathbf{r}(t)=\cos t\,\mathbf{i}+\sin t\,\mathbf{j}+t\,\mathbf{k}\), find \(\mathbf{N}\) and \(\mathbf{B}\), then the normal and osculating planes at \((0,1,\tfrac{\pi}{2})\).

Solution

\(\mathbf{r}'=\langle-\sin t,\cos t,1\rangle\), \(\lvert\mathbf{r}'\rvert=\sqrt2\), so \(\mathbf{T}=\tfrac{1}{\sqrt2}\langle-\sin t,\cos t,1\rangle\).

\(\mathbf{T}'=\tfrac{1}{\sqrt2}\langle-\cos t,-\sin t,0\rangle\), giving \(\mathbf{N}=\langle-\cos t,-\sin t,0\rangle\) — horizontal, pointing at the axis.

\(\mathbf{B}=\mathbf{T}\times\mathbf{N}=\tfrac{1}{\sqrt2}\langle\sin t,-\cos t,1\rangle\).

At \(t=\tfrac{\pi}{2}\): the normal plane has normal \(\langle-2,0,1\rangle\) and the osculating plane reduces to \(z=x+\tfrac{\pi}{2}\).

Admin Note

Radius of curvature \(R=1/\kappa\) is the number that actually shows up in engineering practice — beam bending, lens design, and especially highway/railway geometry. Section 13.4 uses \(\kappa v^{2}\) for the sideways pull on a vehicle; combine it with \(R=1/\kappa\) and you have the standard "design speed vs. curve radius" relationship used to lay out roads.

EXAMPLE 8 Osculating circle of a parabola

Find the osculating circle of \(y=x^{2}\) at the origin.

Solution

For \(y=x^{2}\), \(f'=2x\), \(f''=2\); at the origin \(\kappa=2\), so radius \(R=\tfrac12\).

The circle sits on the concave side with center \(\left(0,\tfrac12\right)\): \(x^{2}+\left(y-\tfrac12\right)^{2}=\tfrac14\).

Section 13.4

Motion in Space: Velocity and Acceleration

Here the calculus of vector functions becomes physics. The derivatives of the position vector are velocity and acceleration, and the section closes by deriving Kepler's First Law from Newton's laws.

Velocity, speed, acceleration

If \(\mathbf{r}(t)\) is position, then velocity is \(\mathbf{v}=\mathbf{r}'\) (tangent to the path), speed is \(\lvert\mathbf{v}\rvert\), and acceleration is \(\mathbf{a}=\mathbf{v}'=\mathbf{r}''\).

Velocity, speed, acceleration

\[ \mathbf{v}(t)=\mathbf{r}'(t),\qquad \text{speed}=\lvert\mathbf{v}(t)\rvert=\frac{ds}{dt},\qquad \mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t) \]

Recovering velocity and position by integration

\[ \mathbf{v}(t)=\mathbf{v}(t_0)+\int_{t_0}^{t}\mathbf{a}(u)\,du,\qquad \mathbf{r}(t)=\mathbf{r}(t_0)+\int_{t_0}^{t}\mathbf{v}(u)\,du \]

EXAMPLE 3 From acceleration to position

A particle starts at \(\mathbf{r}(0)=\langle1,0,0\rangle\) with velocity \(\mathbf{v}(0)=\mathbf{i}-\mathbf{j}+\mathbf{k}\) and has acceleration \(\mathbf{a}(t)=4t\,\mathbf{i}+6t\,\mathbf{j}+\mathbf{k}\). Find \(\mathbf{v}(t)\) and \(\mathbf{r}(t)\).

Solution

Integrate \(\mathbf{a}\): \(\mathbf{v}(t)=2t^{2}\,\mathbf{i}+3t^{2}\,\mathbf{j}+t\,\mathbf{k}+\mathbf{C}\). Using \(\mathbf{v}(0)\), \(\mathbf{C}=\mathbf{i}-\mathbf{j}+\mathbf{k}\).

So \(\mathbf{v}(t)=(2t^{2}+1)\,\mathbf{i}+(3t^{2}-1)\,\mathbf{j}+(t+1)\,\mathbf{k}\).

Integrate again and apply \(\mathbf{r}(0)\): \(\mathbf{r}(t)=\left(\tfrac{2}{3}t^{3}+t+1\right)\mathbf{i}+(t^{3}-t)\,\mathbf{j}+\left(\tfrac12 t^{2}+t\right)\mathbf{k}\).

Newton's Second Law of Motion

If a force \(\mathbf{F}(t)\) acts on an object of mass \(m\) producing acceleration \(\mathbf{a}(t)\), then

\[ \mathbf{F}(t)=m\,\mathbf{a}(t) \]

EXAMPLE 4 Centripetal force

An object of mass \(m\) moves in a circle of radius \(a\) at constant angular speed \(\omega\): \(\mathbf{r}(t)=a\cos\omega t\,\mathbf{i}+a\sin\omega t\,\mathbf{j}\). Find the force.

Solution

\(\mathbf{a}(t)=-\omega^{2}a\cos\omega t\,\mathbf{i}-\omega^{2}a\sin\omega t\,\mathbf{j}=-\omega^{2}\mathbf{r}(t)\).

By Newton's Second Law, \(\mathbf{F}(t)=-m\omega^{2}\mathbf{r}(t)\) — opposite to the radius, pointing at the center. This is centripetal (center-seeking) force.

Projectile motion

Result

With gravity the only force, a projectile launched at angle \(\alpha\) and speed \(v_0\) follows a parabola. The range \(d=\dfrac{v_0^{2}\sin 2\alpha}{g}\) is maximized at \(\alpha=45^\circ\).

Projectile trajectory (launch from origin)

\[ x=(v_0\cos\alpha)\,t,\qquad y=(v_0\sin\alpha)\,t-\tfrac12 g t^{2} \]

EXAMPLE 6 Projectile from a height

A projectile is fired at muzzle speed \(150\ \text{m/s}\), elevation \(45^\circ\), from \(10\ \text{m}\) above the ground. Where does it land?

Solution

With \(g=9.8\), \(v_0=150\), \(\alpha=45^\circ\): \(x=75\sqrt2\,t\), \(\;y=10+75\sqrt2\,t-4.9t^{2}\).

Set \(y=0\) and solve the quadratic; the positive root is \(t\approx21.74\ \text{s}\).

Then \(x=75\sqrt2\,(21.74)\approx2306\ \text{m}\). Impact speed comes from \(\lvert\mathbf{v}\rvert\approx151\ \text{m/s}\).

Tangential and normal components of acceleration

Acceleration always lies in the osculating plane — it has no binormal component. Splitting it along \(\mathbf{T}\) and \(\mathbf{N}\) separates "changing speed" from "changing direction."

Acceleration resolved into components

\[ \mathbf{a}=a_T\,\mathbf{T}+a_N\,\mathbf{N},\qquad a_T=v',\qquad a_N=\kappa v^{2} \]

Computable forms

\[ a_T=\frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{\lvert\mathbf{r}'(t)\rvert},\qquad a_N=\frac{\lvert\mathbf{r}'(t)\times\mathbf{r}''(t)\rvert}{\lvert\mathbf{r}'(t)\rvert} \]

Physical reading

\(a_T=v'\) is the rate of change of speed; \(a_N=\kappa v^{2}\) is the sideways pull. Doubling speed quadruples the normal component — which is why a fast, sharp turn throws a passenger against the door.

EXAMPLE 7 Splitting the acceleration

For \(\mathbf{r}(t)=\langle t^{2},t^{2},t^{3}\rangle\), find the tangential and normal components of acceleration.

Solution

\(\mathbf{r}'=\langle2t,2t,3t^{2}\rangle\), \(\;\mathbf{r}''=\langle2,2,6t\rangle\), \(\;\lvert\mathbf{r}'\rvert=\sqrt{8t^{2}+9t^{4}}\).

\(a_T=\dfrac{\mathbf{r}'\cdot\mathbf{r}''}{\lvert\mathbf{r}'\rvert}=\dfrac{8t+18t^{3}}{\sqrt{8t^{2}+9t^{4}}}\).

\(\mathbf{r}'\times\mathbf{r}''=\langle6t^{2},-6t^{2},0\rangle\), so \(a_N=\dfrac{\lvert\mathbf{r}'\times\mathbf{r}''\rvert}{\lvert\mathbf{r}'\rvert}=\dfrac{6\sqrt2\,t^{2}}{\sqrt{8t^{2}+9t^{4}}}\).

Kepler's laws of planetary motion

The chapter's capstone: using only the calculus of vector functions plus two of Newton's laws, one can derive the empirical laws Kepler extracted from Tycho Brahe's observations.

Kepler's Laws

First. A planet orbits the sun in an ellipse with the sun at one focus.

Second. The line from the sun to a planet sweeps out equal areas in equal times.

Third. The square of a planet's orbital period is proportional to the cube of the major axis length: \(\;T^{2}=\dfrac{4\pi^{2}}{GM}\,a^{3}\).

Newton's Law of Universal Gravitation

\[ \mathbf{F}=-\frac{GMm}{r^{3}}\,\mathbf{r}=-\frac{GMm}{r^{2}}\,\mathbf{u} \]

where \(\mathbf{u}=\mathbf{r}/r\) is the unit radial vector, \(G\) the gravitational constant, and \(M,m\) the masses of sun and planet.

DERIVATION Sketch of Kepler's First Law

Why the orbit is planar and elliptical.

Solution outline

Equate the two expressions for \(\mathbf{a}\): gravity gives \(\mathbf{a}=-\dfrac{GM}{r^{3}}\mathbf{r}\), so \(\mathbf{a}\) is parallel to \(\mathbf{r}\) and \(\mathbf{r}\times\mathbf{a}=\mathbf{0}\).

Then \(\dfrac{d}{dt}(\mathbf{r}\times\mathbf{v})=\mathbf{r}\times\mathbf{a}=\mathbf{0}\), so \(\mathbf{h}=\mathbf{r}\times\mathbf{v}\) is constant. The motion stays in the fixed plane perpendicular to \(\mathbf{h}\).

Computing \(\mathbf{v}\times\mathbf{h}\) and integrating yields \(\mathbf{v}\times\mathbf{h}=GM\,\mathbf{u}+\mathbf{c}\) for a constant vector \(\mathbf{c}\).

Dotting with \(\mathbf{r}\) and converting to polar coordinates produces \(r=\dfrac{ed}{1+e\cos\theta}\) — the polar equation of a conic with eccentricity \(e=\lvert\mathbf{c}\rvert/GM\). Since orbits are closed, the conic is an ellipse.

Admin Note

This derivation is the single best advertisement for the whole chapter: every tool — vector derivatives, the cross-product rule, the constant-length/orthogonality trick from 13.2, and integration of vector functions — is used exactly once to produce a law of nature. If a student asks "where does any of this get used," walk them through these four steps.