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Chapter 14: Problem 53

Verify that the curve \(\mathbf{r}(t)\) lies on the given surface. Give the name of the surface. $$\begin{aligned}&\mathbf{r}(t)=(\sqrt{t^{2}+1} \cos t) \mathbf{i}+(\sqrt{t^{2}+1} \sin t) \mathbf{j}+t \mathbf{k}\\\&x^{2}+y^{2}-z^{2}=1\end{aligned}$$

Short Answer

Expert verified
If yes, what is the type of the surface? Answer: Yes, the curve lies on the surface, and the surface is a hyperboloid of one sheet.

Step by step solution

01

Identify the curve

The given curve \(\mathbf{r}(t)\) has the following components: $$x = \sqrt{t^2+1} \cos t$$ $$y = \sqrt{t^2+1} \sin t$$ $$z = t$$
02

Plug the components into the surface equation

We will now plug the components of the curve into the surface equation \(x^2+y^2-z^2=1\): $$(\sqrt{t^2+1} \cos t)^2 + (\sqrt{t^2+1} \sin t)^2 - t^2 =1$$
03

Simplify the surface equation

Simplify the equation by using the identities \(\sin^2 t + \cos^2 t = 1\) and \((a \cos t)^2 + (a \sin t)^2 = a^2\). $$(t^2+1)(\cos^2 t + \sin^2 t) - t^2 =1$$ $$(t^2+1)(1)-t^2=1$$ $$t^2+1-t^2=1$$
04

Check the validity of the equation

As we can see, both sides of the equation are equal, meaning the curve indeed lies on the surface. $$1=1$$
05

Identify the surface type

The given surface equation is \(x^2+y^2-z^2=1\). This corresponds to the equation of a hyperboloid of one sheet. So the surface is a hyperboloid of one sheet. In conclusion, the curve \(\mathbf{r}(t)\) lies on the surface, and the surface is a hyperboloid of one sheet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations are an efficient way to represent curves and surfaces in mathematics, where each coordinate of a point on the curve or surface is given as a function of one or more parameters.

In the context of our exercise, the parametric equations of the curve are given by:
  • For the x-coordinate, we have:
    \( x(t) = \sqrt{t^2+1} \cos t \)
  • For the y-coordinate, it's:
    \( y(t) = \sqrt{t^2+1} \sin t \)
  • And for the z-coordinate:
    \( z(t) = t \)
This format is particularly helpful for representing the curve's trajectory in a three-dimensional space as a set of functions, where each function corresponds to one of the three dimensions. It's also especially useful when working with curves that do not fit a standard Cartesian equation or for complex motion descriptions in physics and engineering.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. They are essential tools for simplifying expressions and solving equations that involve trigonometric functions.

One of the fundamental identities used in our exercise, which is crucial in simplifying the surface equation, is:
\( \sin^2 t + \cos^2 t = 1 \)
This identity states that the sum of the squares of the sine and cosine of any angle will always equal one. In the verification step of the solution, employing this identity is essential to simplify the expression and establish that the curve indeed lies on the hyperboloid of one sheet. By recognizing and applying such identities, complex algebraic expressions are often transformed into simpler and more solvable forms, making them indispensable in various fields of mathematics and science.
Surface Equation Verification
Surface equation verification is a process used to confirm whether a given set of points, represented by a parametric equation or a vector function, actually resides on a defined surface.

The method involves substituting the parametric equations of the curve into the equation of the surface and simplifying the resultant expression to check for consistency. In our example, the components of the curve defined by \( \mathbf{r}(t) \) are plugged into the equation \( x^2+y^2-z^2=1 \) and simplified using trigonometric identities to verify that the curve lies on the hyperboloid.By confirming that the simplified equation holds true for all values of the parameter, we conclude that the curve is a subset of the surface described by the equation.

Importance of Surface Verification

Surface verification is not only crucial for mathematicians in their theoretical work but also for engineers and physicists when modeling the curves and surfaces in the physical world. It ensures that the parametric equations model the actual shape of the object or trajectory being studied.

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Most popular questions from this chapter

Three-dimensional motion Consider the motion of the following objects. Assume the \(x\) -axis points east, the \(y\) -axis points north, the positive z-axis is vertical and opposite g. the ground is horizontal, and only the gravitational force acts on the object unless otherwise stated. a. Find the velocity and position vectors, for \(t \geq 0\). b. Make a sketch of the trajectory. c. Determine the time of flight and range of the object. d. Determine the maximum height of the object. A soccer ball is kicked from the point \langle 0,0,0\rangle with an initial velocity of \(\langle 0,80,80\rangle \mathrm{ft} / \mathrm{s} .\) The spin on the ball produces an acceleration of \(\langle 1.2,0,0\rangle \mathrm{ft} / \mathrm{s}^{2}\).

Nonuniform straight-line motion Consider the motion of an object given by the position function $$\mathbf{r}(t)=f(t)\langle a, b, c\rangle+\left\langle x_{0}, y_{0}, z_{0}\right\rangle, \quad \text { for } t \geq 0$$ where \(a, b, c, x_{0}, y_{0},\) and \(z_{0}\) are constants, and \(f\) is a differentiable scalar function, for \(t \geq 0\) a. Explain why \(r\) describes motion along a line. b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?

Time of flight, range, height Derive the formulas for time of flight, range, and maximum height in the case that an object is launched from the initial position \(\left\langle 0, y_{0}\right\rangle\) above the horizontal ground with initial velocity \(\left|\mathbf{v}_{0}\right|\langle\cos \alpha, \sin \alpha\rangle\).

Tilted ellipse Consider the curve \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. Assuming the curve lies in a plane, prove that the curve is an ellipse in that plane.

Solving equations of motion Given an acceleration vector, initial velocity \(\left\langle u_{0}, v_{0}, w_{0}\right\rangle,\) and initial position \(\left\langle x_{0}, y_{0}, z_{0}\right\rangle,\) find the velocity and position vectors, for \(t \geq \mathbf{0}\). $$a(t)=\langle 0,0,10\rangle,\left\langle u_{0}, v_{0}, w_{0}\right\rangle=\langle 1,5,0\rangle, \left\langle x_{0}, y_{0}, z_{0}\right\rangle=\langle 0,5,0\rangle$$
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