13.1.1.
What is the domain of \(\v{r}(t) = e^t\v{i} + \dfrac{1}{t}\v{j} + (t + 1)^{-3}\v{k}\text{?}\)
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Worksheet Assigned Problems for Section 13.1
The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
13.1.5.
Find a vector parametrization of the line through \(P = (3,-5,7)\) in the direction \(\v{v} = \la 3,0,1 \ra\text{.}\)
Solution.
We use the vector parametrization of the line to obtain
\begin{align*}
\v{r}(t) \amp= \overrightarrow{OP} + t\v{v} \\
\amp = \la 3, -5, 7 \ra + t\la 3, 0, 1 \ra \\
\amp = \la 3 + 3t, -5, 7 + t \ra
\end{align*}
or in the form of
\begin{equation*}
\v{r}(t) = \lp 3 + 3t \rp \v{i} - 5\v{j} + \lp 7 + t \rp\v{k}
\end{equation*}
13.1.7.
Determine whether the space curve given by \(\v{r}(t) = \la \sin(t), \dfrac{\cos(t)}{2}, t \ra\) intersects the \(z\)-axis, and if it does, determine where.
Solution.
The curve intersects the \(z\)-axis if there is some value of \(t\) such that both the \(x\) and \(y\)-coordinates of \(\v{r}(t)\) are zero. That is, such that \(\sin(t) = \cos\lp\frac{t}{2}\rp = 0\text{.}\) Now, \(\sin(x) = 0\) when \(x = n\pi\text{,}\) while \(\cos(x) = 0\) when \(x = \frac{\pi}{2} + n\pi = \frac{2n+1}{2}\pi\text{,}\) where \(n\) is an integer. So if \(t = k\pi\text{,}\) then \(\sin(t) = 0\text{,}\) and if, further, \(k = 2n + 1\) is odd, then \(\cos\lp\frac{k}{2}\pi\rp = 0\text{.}\) So this curve intersects the \(z\)-axis whenever \(k\) is an odd integer multiple of \(\pi\text{,}\) at the points \(\lp 0,0,(2n + 1)\pi \rp\text{.}\)
13.1.15.
Match the vector-valued functions (a)-(f) with the space curves (i)-(vi) in the figure below.
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\(\displaystyle \v{r}(t) = \la t+15, e^{0.08t}\cos(t), e^{0.08t}\sin(t) \ra\)
-
\(\displaystyle \v{r}(t) = \la \cos(t), \sin(t), \sin(12t) \ra\)
-
\(\displaystyle \v{r}(t) = \la t, t, \dfrac{25t}{1+t^2} \ra\)
-
\(\displaystyle \v{r}(t) = \la \cos^3(t), \sin^3(t), \sin(2t) \ra\)
-
\(\displaystyle \v{r}(t) = \la t, t^2, 2t \ra\)
-
\(\displaystyle \v{r}(t) = \la \cos(t), \sin(t), \cos(t)\sin(12t) \ra\)
Solution.
13.1.17.
Match the space curves (A)-(C) in the figure below with their projections (i)-(iii) onto the \(xy\)-plane.
Solution.
Curve (A) does not cross over itself, and appears to remain a constant distance from the \(z\)-axis, so it corresponds to projection (ii).
Curve (C) oscillates, but does not cross over itself, so it must correspond to projection (i).
Therefore, curve (B) corresponds to projection (iii).
Exercise Group.
In the following exercises, the function \(\v{r}(t)\) traces a circle. Determine the radius, center, and plane containing the circle.
13.1.19.
\(\v{r}(t) = (9\cos(t))\v{i} + (9\sin(t))\v{j}\)
Solution.
13.1.21.
\(\v{r}(t) = \la \sin(t), 0, 4+\cos(t) \ra\)
Solution.
Since \(x(t) = \sin(t)\) and \(z(t) = 4 + \cos(t)\text{,}\) we have
\begin{equation*}
x^2 + (z-4)^2 = \sin^2(t) + \cos^2(t) = 1
\end{equation*}
\(y = 0\) is the equation of the \(xz\)-plane. We conclude that. the function traces the circle of radius \(1\text{,}\) centered at th epoint \((0,0,4)\text{,}\) and contained in the \(xz\)-plane.
13.1.25.
Let \(\c{C}\) be the curve given by \(\v{r}(t) = \la t\cos(t), t\sin(t), t \ra\text{.}\)
-
Sketch the cone and make a rough sketch of \(\c{C}\) on the cone.
Solution.
-
We have \(x(t) = t\cos(t)\text{,}\) \(y(t) = t\sin(t)\text{,}\) and \(z(t) = t\text{.}\) Hence for any point \((x,y,z)\) on the curve, we obtain\begin{align*} x^2 + y^2 \amp= t^2 \cos^2(t) + t^2 \sin^2(t) \\ \amp= t^2 \lp \cos^2(t) + \sin^2(t) \rp \\ \amp= t^2 \\ \amp= z^2 \end{align*}Thus the curve lies on the circular cone whose equation is \(x^2 + y^2 = z^2\text{.}\)
-
As the height \(z = t\) increases linearly with time, the \(x\) and \(y\) coordinates trace out points on the circles of increasing radius. We obtain the following curve.
Figure 13.1.13. The Sketch of the Curve \(\v{r}(t) = \la t\cos(t), t\sin(t), t \ra\text{.}\)
13.1.29.
Parametrize the part of the intersection of the surfaces
\begin{equation*}
y^2 - z^2 = x - 2\, , \qquad y^2 + z^2 = 9
\end{equation*}
where \(z \geq 0\) using \(t = y\) as the parameter.
Solution.
We solve for \(z\) and \(x\) in terms of \(y\text{.}\) From the equation \(y^2 + z^2 = 9\text{,}\) we have \(z^2 = 9 - y^2\text{,}\) so \(z = \pm \sqrt{9 - y^2}\text{.}\) From the second equation, we have
\begin{align*}
x \amp= y^2 - z^2 + 2 \\
\amp= y^2 - \lp 9 - y^2 \rp + 2 \\
\amp= 2y^2 - 7
\end{align*}
Taking \(t = y\) as a parameter, we have
\begin{equation*}
z = \pm \sqrt{9 - t^2} \qquad \text{ and } \qquad x = 2t^2 - 7
\end{equation*}
yielding the following vector parametrization
\begin{equation*}
\v{r}(t) = \la 2t^2 - 7, t, \pm \sqrt{9 - t^2} \ra \, , \quad \text{for } -3 \leq t \leq 3
\end{equation*}
13.1.33.
Use sine and cosine to parametrize the intersection of the cylinder \(x^2 + y^2 = 1\) and the plane \(x + y + z = 1\text{.}\) Then describe the projections of this curve onto the three coordinate planes.
Solution.
To parametrize the intersection, we start with the cylinder equation \(x^2 + y^2 = 1\text{.}\) This suggests using polar coordinates for \(x\) and \(y\text{.}\) Let \(x = \cos(t)\) and \(y = \sin(t)\) for \(0 \leq t \leq 2\pi\text{.}\)
We find the \(z\) component by substituting these into the equation of the plane \(x + y + z = 1\text{:}\)
\begin{equation*}
z = 1 - x - y = 1 - \cos(t) - \sin(t)
\end{equation*}
Thus, the vector parametrization is:
\begin{equation*}
\v{r}(t) = \la \cos(t), \sin(t), 1 - \cos(t) - \sin(t) \ra
\end{equation*}
The projections onto the coordinate planes are found by eliminating the variable corresponding to the axis perpendicular to that plane:
-
The \(xy\)-plane (\(z=0\)).The projection is the unit circle \(x^2 + y^2 = 1\text{.}\)
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The \(xz\)-plane (\(y=0\)).We eliminate \(y\) using the plane equation \(y = 1 - x - z\) and substitute it into the cylinder equation. The projection is the ellipse given by \(x^2 + (1 - x - z)^2 = 1\text{.}\)
-
The \(yz\)-plane (\(x=0\)).We eliminate \(x\) using the plane equation \(x = 1 - y - z\) and substitute it into the cylinder equation. The projection is the ellipse given by \((1 - y - z)^2 + y^2 = 1\text{.}\)
13.1.37.
Two paths \(\v{r}_1(t)\) and \(\v{r}_2(t)\) intersect if there is a point \(P\) lying on both curves. We say that \(\v{r}_1(t)\) and \(\v{r}_2(t)\) collide if \(\v{r}_1(t_0) = \v{r}_2(t_0)\) at some point \(t_0\text{.}\)
Determine whether \(\v{r}_1(t)\) and \(\v{r}_2(t)\) collide or intersect, giving the coordinates of the corresponding points if they exist:
\begin{equation*}
\v{r}_1(t) = \la t^3+3, t+1, 6t^{-1} \ra\, , \qquad \v{r}_2(t) = \la 4t, 2t-2, t^2-7 \ra
\end{equation*}
Solution.
Collision: We check if there is a time \(t\) such that \(\v{r}_1(t) = \v{r}_2(t)\text{.}\) Equating the components:
\begin{align*}
t^3 + 3 \amp= 4t \implies t^3 - 4t + 3 = 0\\
t + 1 \amp= 2t - 2 \implies t = 3
\end{align*}
For the \(y\)-components to match, we must have \(t=3\text{.}\) However, testing \(t=3\) in the \(x\)-equation gives \((3)^3 - 4(3) + 3 = 27 - 12 + 3 = 18 \neq 0\text{.}\) Since no single value of \(t\) satisfies all equations, the particles do not collide.
Intersection: We check if the curves share a point by using different parameters \(t\) and \(s\text{.}\) We set \(\v{r}_1(t) = \v{r}_2(s)\text{:}\)
\begin{align*}
1) \quad t^3 + 3 \amp= 4s\\
2) \quad t + 1 \amp= 2s - 2\\
3) \quad \frac{6}{t} \amp= s^2 - 7
\end{align*}
From equation (2), we solve for \(s\text{:}\) \(2s = t + 3 \implies s = \frac{t+3}{2}\text{.}\) Substituting this into equation (1):
\begin{equation*}
t^3 + 3 = 4\lp\frac{t+3}{2}\rp = 2(t+3) = 2t + 6
\end{equation*}
\begin{equation*}
t^3 - 2t - 3 = 0
\end{equation*}
Substituting \(s = \frac{t+3}{2}\) into equation (3):
\begin{equation*}
\frac{6}{t} = \lp\frac{t+3}{2}\rp^2 - 7 \implies \frac{24}{t} = (t+3)^2 - 28
\end{equation*}
Simplifying this yields \(t^3 + 6t^2 - 19t - 24 = 0\text{.}\) We check the integer roots of this polynomial (\(t = -1, 3, -8\)) to see if they satisfy our condition from equation (1), \(t^3 - 2t - 3 = 0\text{.}\)
\begin{equation*}
\text{If } t = -1: \quad (-1)^3 - 2(-1) - 3 = -2 \neq 0
\end{equation*}
\begin{equation*}
\text{If } t = 3: \quad (3)^3 - 2(3) - 3 = 18 \neq 0
\end{equation*}
As there is no value \(t\) that satisfies the system for all three coordinates, the paths do not intersect.
Exercise Group.
In the following exercises, find a parametrization of the curve.
13.1.43.
Solution.
The circle is parallel to the \(yz\)-plane and centered at \((1,2,5)\text{,}\) hence the \(x\)-coordinates of the points on the circle are \(x = 1\text{.}\) The projection of the circle on the \(yz\)-plane is a circle of radius \(2\) centered at \((2,5)\text{.}\) This circle is parametrized by
\begin{equation*}
y = 2 + 2\cos(t) \qquad \text{ and } \qquad z = 5 + 2\sin(t)
\end{equation*}
We conclude that the points on the required circle can be written as \(\lp 1, 2 + 2\cos(t), 5 + 2\sin(t) \rp\text{.}\) This gives the following parametrization
\begin{equation*}
\v{r}(t) = \la 1, 2+2\cos(t), 5+2\sin(t) \ra
\end{equation*}
13.1.45.
The intersection of the plane \(y = \dfrac{1}{2}\) with the sphere \(x^2 + y^2 + z^2 = 1\text{.}\)
Solution.
Substituting \(y = \frac{1}{2}\) in the equation of the sphere, we obtain
\begin{equation*}
x^2 + \lp \frac{1}{2} \rp^2 + z^2 = 1
\qquad \implies \qquad
x^2 + z^2 = \frac{3}{4}
\end{equation*}
This circle in the horizontal plane \(y = \frac{1}{2}\) has the parametrization
\begin{equation*}
x = \frac{\sqrt{3}}{2} \cos(t) \qquad \text{ and } \qquad z = \frac{\sqrt{3}}{2} \sin(t)
\end{equation*}
Therefore, the points on the intersection of the plane \(y = \frac{1}{2}\) and the sphere \(x^2 + y^2 + z^2 = 1\) can be written in the form \(\lp \frac{\sqrt{3}}{2}\cos(t), \frac{1}{2}, \frac{\sqrt{3}}{2}\sin(t) \rp\text{,}\) yielding the following parametrization
\begin{equation*}
\v{r}(t) = \la \frac{\sqrt{3}}{2}\cos(t), \frac{1}{2}, \frac{\sqrt{3}}{2}\sin(t) \ra
\end{equation*}
