We again remind you to read through the introduction to this chapter 10, which introduces the Classic Calculus Approach and reviews the high points of single variable calculus. You may want to have students read this review of single variable calculus so they can recall the various uses of limits, derivatives, and integrals before we start working with these concepts in more general settings.
The material in this section requires students to understand the graph of a vector valued function of one variable (from Section 10.1) and our vector operations, like addition and scalar multiplication, from Section 9.2.
In this book, we will use the structure of the classic calculus approach to investigate properties of new kinds of functions, equations, and graphs. Later in this section, we will investigate the meaning of limits, derivatives, and integrals of vector-valued functions of one variable. Additionally, we will connect these measurements to properties of curves in space as the graphs of these vector-valued functions.
As the only employee of Steer Clear, you have decided that you need to understand how the timing of position measurements will change different properties related to your self-driving car. You decide to drive in a figure eight path described by \(\vr_8(t)=\langle \cos(t),\sin(2t) \rangle \) for \(0 \leq t \leq 2 \pi\text{.}\) A plot of this path is given in Figure 10.2.1.
A curve in the plane. The curve intersects itself at the origin and is symmetric about both axes. On each side of the vertical axis, the curve forms a sort of squashed loop with a point at the origin.
In order to understand how often your software should collect location data, you decide to look at your position for a few different times. Calculate the following, rounding the component values to three decimal places. Draw the output vectors of \(\vr_8\) on Figure 10.2.1 in standard position.
The tips of these vectors correspond to the locations that would be sampled if you wanted to know the location of your car at \(\vr_8(\pi)\) but collected data every second, every tenth of a second, and every hundredth of a second, respectively.
Calculate \(\vr_8(\pi)\) and \(\vr_8(\pi)-\vr_8(3)\text{.}\) Sketch \(\vr_8(\pi)\) and \(\vr_8(3)\) in standard position in Figure 10.2.1. Also plot \(\vr_8(\pi)-\vr_8(3)\text{,}\) positioned in such a way to illustrate that it is difference of these two position vectors.
How would you expect \(\frac{\vr_8(\pi)-\vr_8(3.1)}{\pi-3.1}\) to be different than \(\frac{\vr_8(\pi)-\vr_8(3)}{\pi-3}\text{?}\) Use this idea write about what is measured by \(\frac{\vr_8(\pi)-\vr_8(\pi-h)}{h}\) if we look at smaller and smaller values of \(h\text{.}\) Remember to be specific about what aspects of our curve or drive are being measured.
In part b of the Preview Activity, we examined what the limit of a vector-valued function will look like. Specifically, the limit of a vector-valued function of one variable has the same meaning as in your first calculus course. The limit of a function is a measure of what the output approaches as the input gets closer to a particular value. We will not state the formal version of limits of vector-valued functions of one variable, but rather will appeal to the important idea that limits can be defined and calculated componentwise. If \(\vr(t)=\langle x_1(t),\ldots,x_n(t)\rangle\text{,}\) then the limit of \(\vr\) as \(t\rightarrow a\) exists if all of the limits of \(x_i(t)\) exist as \(t\rightarrow a\text{.}\) In other words, the limit of a vector-valued function of one variable is the same as the limits of the \(n\) component functions organized as a vector. In \(\mathbb{R}^3\text{,}\) we can express this algebraically as
provided the limit exists. At a given value of \(a\text{,}\)\(f'(a)\) measures the instantaneous rate of change of \(f\text{,}\) and also tells us the slope of the tangent line to the curve \(y = f(x)\) at the point \((a, f(a))\text{.}\) The definition of the derivative extends naturally to vector-valued functions and curves in space.
for those values of \(t\) at which the limit exists. We also use the notation \(\frac{d\vr}{dt}\) and \(\frac{d}{dt}[\vr(t)]\) for \(\vr\,'(t)\text{.}\)
We will take a moment here to note that both the single variable calculus definition of derivative and Definition 10.2.2 have the form of the classic calculus approach described above. In particular, both definitions are of the form
\begin{equation*}
\text{derivative}=\lim_{\text{stepsize}\rightarrow 0} \frac{\text{change in output of function}}{\text{change in input of function}}
\end{equation*}
Let \(\vr\) be the vector-valued function whose graph is shown in Figure 10.2.3, and let \(h\) be a scalar that represents a small change in time. The vector \(\vr(t)\) is the blue vector in Figure 10.2.3 and \(\vr(t+h)\) is the green vector.
If \(\vr(t)\) determines the position of an object at time \(t\text{,}\) then \(\frac{\vr(t+h)-\vr(t)}{h}\) represents the average rate of change in the position of the object over the interval \([t,t+h]\text{.}\) We also call this the average velocity of the object on the interval\([t,t+h]\text{.}\) It may feel a bit surprising to think of a rate of change as a vector quantity. However, describing how the vector quantity position changes with respect to time requires knowing both about the direction in which position changes and the magnitude of that change. Furthermore, when computing the difference quotient, the vector subtraction in the numerator is done componentwise. The factor of \(1/h\) is also multiplied into each component, so we can view the difference quotient’s components as each being the average rate of change for that particular component.
is the instantaneous rate of change of \(\vr(t)\) at time \(t\) for those values of \(t\) for which the limit exists. Therefore, the instantaneous velocity of the object at time \(t\) is \(\vv(t) = \vr\, '(t)\text{.}\)Figure 10.2.4 allows you to reduce the step size in the difference quotient to approach \(0\text{.}\) In doing so, you should notice that \(\vr\, '(t)\) is tangent to the graph of \(\vr(t)\) at time \(t\text{.}\) We will return to this important characteristic later in the section.
is the instantaneous rate of change of the velocity of the object at time \(t\text{,}\) for those values of \(t\) for which the limits exists. Therefore, the acceleration of the moving object is given by \(\va(t) = \vv\, '(t)\text{.}\)
Both the velocity and acceleration are vector quantities: they have magnitude and direction. In single-variable calculus, we distinguish the velocity of an object, which can be positive, negative, or zero, from its speed, which must be nonnegative. The speed of an object in the single-variable case is the absolute value of its velocity. The analogous idea for an object having movement described by a vector-valued function is that the magnitude of the object’s velocity vector is the speed of the object. This allows the speed \(\vecmag{\vv(t)}\) to again be a nonnegative scalar quantity.
As we learned in single variable calculus, computing derivatives using the definition is often difficult because the limit involved is an indeterminate form. Fortunately, properties of limits and vectors make it straightforward to calculate the derivative of a vector-valued function. This will be similar to how we developed shortcut differentiation rules in single-variable calculus. If we analyze the steps in Definition 10.2.2, we identify three steps
each of these is all evaluated componentwise on a vector-valued function. Therefore, familiar differentiation rules from single-variable calculus can be applied one ach component of \(\vr(t)\) to compute the derivative \(\vr\, '\text{.}\)
To see why, recall that the limit of a sum is the sum of the limits, and that we can remove constant factors from limits. Thus, as we observed after Preview Activity 10.2.1, if \(\vr(t) = x(t) \vi + y(t) \vj + z(t) \vk\text{,}\) it follows that
Our next activity gives you a chance to practice with componentwise differentiation and also asks you to think about when limits of vector-valued functions do not exist.
For each of the following vector-valued functions, state any values of \(a\) for which \(\displaystyle\lim_{t\to a}\vr(t)\) will not exist and find \(\vr\, '(t)\text{.}\)
In first-semester calculus, we developed several important differentiation rules, including the constant multiple, product, quotient, and chain rules. For instance, we can formally state the product rule as
There are several analogous rules for vector-valued functions, including a product rule for each of the products we have defined involving vectors. These rules, which are easily verified by componentwise application of the familiar single-variable calculus rules, are summarized below.
Properties of Derivatives of Vector-Valued Functions.
Let \(f\) be a differentiable real-valued function of a real variable \(t\text{.}\) Let \(\vr\) and \(\vs\) be differentiable vector-valued functions of the real variable \(t\text{.}\) Then
When applying these properties, pay attention to whether the quantities involved are either scalars or vectors. For example, \(\vr(t) \cdot \vs(t)\) defines a scalar function because we have taken the dot product of two vector-valued functions. However, \(\vr(t) \times \vs(t)\) defines a vector-valued function since we have taken the cross product of two vector-valued functions. Also notice that while we have three product rules for the differentiation of vector-valued functions, we have only one chain rule and no quotient rule. The lack of a quotient rule should make sense, since we have not defined what it would mean to divide a vector by another. That we have a single chain rule arises because so far, we have only discussed how to have a single input to a vector-valued function. Thus, we cannot compose one vector-valued function with another vector-valued function. You might also notice that the order in which the factors in the Chain Rule are written is a bit different to what you would write for single-variable functions. This is because \(f'(t)\) is a scalar, and convention says we should write scalar-vector multiplication with the scalar first.
In the next several sections of this chapter, we will examine how to use calculus tools and vector measurements to calculate different properties of curves in space and the associated velocity and acceleration vectors that come from their parameterizations. Exercise 14 offers practice on using calculus tools on vector-valued functions and previews some ideas that will be used in later sections of this chapter.
One of the most important ideas in first-semester calculus is that a differentiable function is locally linear: that is, when viewed up close, the graph of a differentiable function looks like a line. Indeed, when we zoom in sufficiently close to a particular point, the curve is indistinguishable from its tangent line. You can look at Figure 9.5.1 for an interactive visualization of this important idea.
In the same way, we expect that a smooth curve in three-dimensional space will be locally linear if the graph of the curve looks like a line when you zoom in sufficently far. In Activity 10.2.3, we investigate how to find the tangent line to such a curve. Recall from our work in Section 9.5 that the vector equation of a line that passes through the point at the tip of the vector \(\vL_0 = \langle x_0, y_0, z_0 \rangle\) in the direction of the vector \(\vu = \langle a, b, c \rangle\) can be written as
\begin{equation*}
\vL(t) = \vL_0 + t \vu\text{.}
\end{equation*}
The line described by \(\vL\) can also be written in parametric form as
Figure 10.2.6 shows a plot of a curve, a point on that curve, and the line tangent to the curve at that point. You can change the point at which the tangent line is shown by moving the “Location” slider. By moving the “Zoom” slider to the right, you can show smaller and smaller regions around the highlighted point to see that the tangent line approximates the curve very well on sufficiently small neighborhoods of the point. If you move the zoom slider all the way to the left, the whole curve and axes will be shown.
Recall that we discussed earlier that the vector \(\vr\, '(a)\) is tangent to the graph of \(\vr(t)\) at the point where \(t=a\text{.}\) Find a direction vector for the line tangent to the graph of \(\vr\) at the point where \(t=\pi\text{.}\)
On your plot of the curve \(\vr(t)\text{,}\) sketch the tangent line corresponding to \(t = \pi\) and highlight the role of \(\vr\, '(\pi)\) on your plot.
Our work in Activity 10.2.3 can be generalized to give a method for finding an equation of the line tangent to a curve described by a vector-valued function.
The tangent line is determined entirely by a given point (provided by the function \(\vr\)) and a direction vector (provided by the derivative, \(\vr\, '\)), with both the function and its derivative evaluated at \(t = a\text{.}\) Note how analogous the formula for \(\vL(t)\) is to the tangent line approximation from single-variable calculus: in that context, for a given function \(y = f(x)\) at a value \(x = a\text{,}\) we found that the tangent line can be expressed by the linear function \(y = L(x)\) whose formula is
Equation (10.2.1) for the tangent line \(\vL(t)\) to the vector-valued function \(\vr(t)\) is nearly identical. Indeed, because there are multiple parameterizations for a single line, it is even possible to write the parameterization as
For example, in Equation (10.2.1), \(\vL(0) = \vr(a)\text{,}\) so the line’s parameterization “starts” at \(t = 0\text{.}\) When we write the parameterization in the form of Equation (10.2.2), \(\vL(a) = \vr(a)\text{,}\) so the line’s parameterization “starts” at \(t = a\text{.}\) Both of these ways of parameterizing the line tangent to the graph of a vector-valued function will be useful in our future work.
where \(\Delta x_i\) is the width of the interval from \([x_{i-1},x_i]\) and \(x^*_i\) is a point in \([x_{i-1},x_i]\text{.}\) The definite integral measures the net signed area between the graph of \(f(x)\) and the \(x\)-axis on the interval from \(x=a\) to \(x=b\text{.}\) The definite integral uses our classic calculus approach to measure the accumulation of the output of \(f\) over the interval \([a,b]\text{.}\) In practice, you probably did not use this Riemann sum definition to calculate any definite integrals, since the Fundamental Therom of Calculus allowed you to use associated antiderivatives to efficiently calculate the value of a definite integral. The Fundamental Theorem of Integral Calculus states that if \(f\) is a continuous function on \([a,b]\) and \(F(x)\) is an antiderivatve of \(f\) (that is, \(\frac{dF}{dx}=f(x)\)), then
The family of functions that are antiderivatives of \(f(x)\) is also called the indefinite integral of \(f(x)\) and is denoted \(\int f(x) \ dx\text{.}\)
We can apply all of the ideas above related to definite and indefinite integrals to vector-valued functions of one variable by performing each operation separately on each component.
provided that all three of the integrals on the right-hand side exist. An antiderivative of a vector-valued function \(\vr\) is a vector-valued function \(\vR\) such that
The indefinite integral \(\int \vr(t) \ dt\) of a vector-valued function \(\vr\) is the most general antiderivative of \(\vr\) and represents the family of all antiderivatives of \(\vr\text{.}\) The indefinite integral of \(\vr(t)\) is denoted by
The area under the curve idea does not generalize when we think about the integral of \(\vr(t)\text{.}\) One way to understand this is that the integral of \(\vr\) is with respect to the input variable \(t\) and there is not a \(t\)-axis on the graph of \(\vr(t)\text{.}\) However, as described in the introduction to this section, the concept that will generalize to vector-valued functions is the definite integral as a measurement of the accumulation of the output of a function. Specifically, in Recovering Position from Velocity, we stated how we can recover the position function by integrating the velocity function. This is a way of viewing position as the accumulation of the rate of change of position. If we apply our argument for the accumulation of rates of change to each component separately, then we can integrate the velocity componentwise to obtain the components of the position vector separately. Additionally, we can integrate acceleration to get the velocity as a function of \(t\text{.}\) Whether we are integrating velocity to get position or acceleration to get velocity, it will be necessary to have a particular point (in position or velocity) that will allow us to solve for the particular integration constant (the “\(+C\)” from single-variable calculus).
For this example, we consider the motion of a projectile in two dimensions where the only force on the projectile is gravity. This type of problem applies to sports like archery or shotput as well as to military applications like mortar or artillery placements. It also has applications to firefighting or fireworks construction. This kind of simplified physics problem is common in introductory courses because there is only one force involved, the force is constant (does not change based on location, speed, or time), and the force is in only one coordinate direction. Even this description is a greatly simplified version of how gravity acts in a greater sense.
Newton’s second law of motion says that that sum of the forces acting on an object is equal to the mass of the object times the acceleration vector. In our problem, the only force is gravity pulling the object in the negative \(y\)-direction. We want to completely determine the path traveled by an object that is launched from a fixed position at a given angle from the horizontal with a given initial velocity. This information can be stated as follows with the initial values shown in Figure 10.2.8:
The acceleration is given by \(m \va(t) = -mg \vj\) (the only force is gravity pulling down with a constant force).
We can simplify \(m \va(t) = -mg \vj\) to find \(\va(t)=\langle 0,-g\rangle\text{.}\) Next, we integrate the acceleration to find the velocity as a function of \(t\text{.}\) The indefinite integral of \(\va(t)=\langle 0,-g\rangle\) with respect to \(t\) will be \(\langle 0,-gt\rangle +\vec{C}\text{,}\) where \(\vec{C}\) is a constant vector. We need to pick the constant such that the initial velocity is \(\vv(0)=\langle v_0 \cos(\theta),v_0 \sin(\theta) \rangle\text{.}\) This will ensure that the initial speed is \(v_0\) and that the initial velocity has an angle of \(\theta\) above the horizontal. Because \(\vv(0)=\langle 0,-g(0)\rangle +\vec{C}=\vec{C}=\langle v_0 \cos(\theta),v_0 \sin(\theta) \rangle\text{,}\) we find the following for the velocity as a function of time:
Now we can integrate our velocity function to get the position as a function of \(t\text{.}\) The indefinite integral of \(\langle v_0 \cos(\theta),v_0 \sin(\theta)-gt \rangle\) will be
Again we need to find the integration constant \(\vec{D}\) that satisifes the initial position \(\vr(0)=\langle x_0,y_0\rangle \text{.}\) This gives the following vector-valued function for the position of the object as a function of time.
In the next activity, we look at relationships between position, velocity, and acceleration in a situation that is more complicated than the constant acceleration of Example 10.2.7. The process of using integration and differentiation will remain the same, but the results will not be nearly as simple.
A graph of the position of the object for times \(t\) in \([-0.5,3]\) is shown in Figure 10.2.9. Suppose further that the object is at the point \((1.5,-1,0)\) at time \(t=0\text{.}\)
The previous activity shows a case where the motion of our object is driven by forces that either turn the object or change the speed of the object in the \(z\)-coordinate direction. We will spend the rest of this chapter looking at measurements related to motion along curves like this. Key to these measurements will be the combined use of our new calculus for vector-valued functions with the vector tools from the previous chapter.
The derivative \(\vr\, '(t)\) of the vector-valued function \(\vr\) tells us the instantaneous rate of change of \(\vr\) with respect to time, \(t\text{,}\) which can be interpreted as a direction vector for the line tangent to the graph of \(\vr\) at the point \(\vr(t)\text{,}\) or also as the instantaneous velocity of an object traveling along the graph defined by \(\vr(t)\) at time \(t\text{.}\)
An antiderivative of \(\vr\) is a vector-valued function \(\vR\) such that \(\vR'(t) = \vr(t)\text{.}\) The indefinite integral \(\int \vr(t) \ dt\) of a vector-valued function \(\vr\) is the general antiderivative of \(\vr\) (which is a collection of all of the antiderivatives of \(\vr\text{,}\) with any two antiderivatives differing by at most a constant vector). Moreover, if \(\vr(t) = x(t) \vi + y(t) \vj + z(t) \vk\text{,}\) then
If an object is launched from a point \((x_0,y_0)\) with initial velocity \(v_0\) at an angle \(\theta\) with the horizontal, then the position of the object at time \(t\) is given by
The WeBWorK problems are written by many different authors. Some authors use parentheses when writing vectors, e.g., \((x(t),y(t),z(t))\) instead of angle brackets \(\langle x(t),y(t),z(t) \rangle\text{.}\) Please keep this in mind when working WeBWorK exercises.
Given that the acceleration vector is \(\mathbf{a} \left( t \right) = \left(
-16 \cos \left( -4 t \right) \right) \mathbf{i} + \left( -16 \sin
\left( -4 t \right) \right) \mathbf{j} + \left( -2 t \right) \mathbf{k}\text{,}\) the initial velocity is \(\mathbf{v} \left( 0 \right) = \mathbf{i + k}\text{,}\) and the initial position vector is \(\mathbf{r} \left( 0 \right) = \mathbf{i +
j + k}\text{,}\) compute:
Suppose \(\vec{r}(t) = \cos\!\left(\pi t\right)\boldsymbol{i}+\sin\!\left(\pi t\right)\boldsymbol{j}+2t\boldsymbol{k}\) represents the position of a particle on a helix, where \(z\) is the height of the particle.
(c) When the particle has height \(8\text{,}\) it leaves the helix and moves along the tangent line at the constant velocity found in part (b). Find a vector parametric equation for the position of the particle (in terms of the original parameter \(t\)) as it moves along this tangent line.
A gun has a muzzle speed of 80 meters per second. What angle of elevation should be used to hit an object 170 meters away? Neglect air resistance and use \(g
= 9.8\, \textrm{m}/\textrm{sec}^{2}\) as the acceleration of gravity.
A child wanders slowly down a circular staircase from the top of a tower. With \(x,y,z\) in feet and the origin at the base of the tower, her position \(t\) minutes from the start is given by
\begin{equation*}
x=25\cos t,\quad y = 25 \sin t, \quad z = 120 - 5 t.
\end{equation*}
Compute the derivative of each of the following functions in two different ways: (1) use the rules provided in the theorem stated just after Activity 10.2.2, and (2) rewrite each given function so that it is stated as a single function (either a scalar function or a vector-valued function with three components), and differentiate component-wise. Compare your answers to ensure that they are the same.
Indicate on the left of Figure 10.2.10 the object’s position, velocity and acceleration at the times \(t=0, 2, 4\text{.}\) Draw the velocity and acceleration vectors with their tails placed at the object’s position.
Recall that the speed is \(\vecmag{\vv} = \sqrt{\vv\cdot\vv}\text{.}\) Find the object’s speed and graph it as a function of time \(t\) on the right of Figure 10.2.10. When is the object’s speed the slowest? When is the speed increasing? When is it decreasing?
What seems to be true about the angle between \(\vv\) and \(\va\) when the speed is at a minimum? What is the angle between \(\vv\) and \(\va\) when the speed is increasing? when the speed is decreasing?
Since the square root is an increasing function, we see that the speed increases precisely when \(\vv\cdot\vv\) is increasing. Use the product rule for the dot product to express \(\frac{d}{dt}(\vv\cdot\vv)\) in terms of the velocity \(\vv\) and acceleration \(\va\text{.}\) Use this to explain why the speed is increasing when \(\vv\cdot\va \gt 0\) and decreasing when \(\vv\cdot\va \lt 0\text{.}\) Compare this to part d.
\begin{equation*}
\vw(s) = \left\langle s^2, \sin\left(\frac{\pi}{2}s\right), s \right\rangle.
\end{equation*}
Determine the point of intersection of the curves generated by \(\vr(t)\) and \(\vw(s)\text{.}\) To do so, you will have to find values of \(a\) and \(b\) that result in \(\vr(a)\) and \(\vw(b)\) being the same vector.
Suppose that \(z = f(x,y)\) is a function that generates a surface in three-dimensional space, and that the curves generated by \(\vr(t)\) and \(\vw(s)\) both lie on this surface. Note particularly that the point of intersection you found in (a) lies on this surface. In addition, observe that the two tangent lines found in (b) and (c) both lie in the tangent plane to the surface at the point of intersection. Use your preceding work to determine the equation of this tangent plane.
For each given function \(\vr\text{,}\) determine \(\int \vr(t) \ dt\text{.}\) In addition, recalling the Fundamental Theorem of Calculus for functions of a single variable, also evaluate \(\int_0^1 \vr(t) \ dt\) for each given function \(\vr\text{.}\) Is the resulting quantity a scalar or a vector? What does it measure?
In this exercise, we develop the formula for the position function of a projectile that has been launched at an initial speed of \(|\vv_0|\) and a launch angle of \(\theta.\) Recall that \(\va(t) = \langle 0, -g \rangle\) is the constant acceleration of the projectile at any time \(t\text{.}\)
Find all velocity vectors for the given acceleration vector \(\va\text{.}\) When you anti-differentiate, remember that there is an arbitrary constant that arises in each component.
Use the given information about initial speed and launch angle to find \(\vv_0\text{,}\) the initial velocity of the projectile. You will want to write the vector in terms of its components, which will involve \(\sin(\theta)\) and \(\cos(\theta)\text{.}\)
Next, find the specific velocity vector function \(\vv\) for the projectile. That is, combine your work in (a) and (b) in order to determine expressions in terms of \(|\vv_0|\) and \(\theta\) for the constants that arose when integrating.
Let \(\vr(t)\) denote the position vector function for the given projectile. Use the fact that the object is fired from the position \((x_0, y_0)\) to show it follows that
A central force is one that acts on an object so that the force \(\vF\) is parallel to the object’s position \(\vr\text{.}\) Since Newton’s Second Law says that an object’s acceleration is proportional to the force exerted on it, the acceleration \(\va\) of an object moving under a central force will be parallel to its position \(\vr\text{.}\) For instance, the Earth’s acceleration due to the gravitational force that the sun exerts on the Earth is parallel to the Earth’s position vector as shown in Figure 10.2.11.
If an object of mass \(m\) is moving under a central force, the angular momentum vector is defined to be \(\vL=m\vr\times\vv\text{.}\) Assuming the mass is constant, show that the angular momentum is constant by showing that
