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Chapter 12: Problem 50
Use cross products to determine whether the points \(A, B,\) and C are collinear. $$A(3,2,1), B(5,4,7), \text { and } C(9,8,19)$$
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Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\) the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C\) the outward-pointing unit vector normal to \(C\) is also $$ \mathbf{n}=\langle\cos \theta, \sin \theta\rangle $$ c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\) find the component of \(v\) normal to \(C\) Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?
For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle-1,2,3\rangle, \mathbf{v}=\langle 2,1,1\rangle\)
In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.
Parabolic trajectory Consider the parabolic trajectory $$ x=\left(V_{0} \cos \alpha\right) t, y=\left(V_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} $$ where \(V_{0}\) is the initial speed, \(\alpha\) is the angle of launch, and \(g\) is the acceleration due to gravity. Consider all times \([0, T]\) for which \(y \geq 0\) a. Find and graph the speed, for \(0 \leq t \leq T.\) b. Find and graph the curvature, for \(0 \leq t \leq T.\) c. At what times (if any) do the speed and curvature have maximum and minimum values?
Let \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\). a. Assume that \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle,\) which means that $$\begin{aligned} &\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0 . \text { Prove that }\\\ &\lim _{t \rightarrow a} f(t)=L_{1}, \quad \lim _{t \rightarrow a} g(t)=L_{2}, \text { and } \lim _{t \rightarrow a} h(t)=L_{3}. \end{aligned}$$ $$\begin{aligned} &\text { b. Assume that } \lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text { and }\\\ &\lim _{t \rightarrow a} h(t)=L_{3} . \text { Prove that } \lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3},\right\rangle\\\ &\text { which means that } \lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0. \end{aligned}$$
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