Chapter 11: Problem 58
Find the domain of the following vector-valued functions. $$\mathbf{r}(t)=\cos 2 t \mathbf{i}+e^{\sqrt{t}} \mathbf{j}+\frac{12}{t} \mathbf{k}$$
Chapter 11: Problem 58
Find the domain of the following vector-valued functions. $$\mathbf{r}(t)=\cos 2 t \mathbf{i}+e^{\sqrt{t}} \mathbf{j}+\frac{12}{t} \mathbf{k}$$
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Get started for freeDistance between a point and a line in the plane Use projections to find a general formula for the (least) distance between the point \(\left.P\left(x_{0}, y_{0}\right) \text { and the line } a x+b y=c . \text { (See Exercises } 62-65 .\right)\).
Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.
Let $$\mathbf{u}(t)=2 t^{3} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}-8 \mathbf{k} \text { and } \mathbf{v}(t)=e^{t} \mathbf{i}+2 e^{-t} \mathbf{j}-e^{2 t} \mathbf{k}$$ Compute the derivative of the following functions. $$\mathbf{u}(t) \cdot \mathbf{v}(t)$$
Direction angles and cosines Let \(\mathbf{v}=\langle a, b, c\rangle\) and let \(\alpha, \beta\) and \(\gamma\) be the angles between \(\mathbf{v}\) and the positive \(x\) -axis, the positive \(y\) -axis, and the positive \(z\) -axis, respectively (see figure). a. Prove that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) b. Find a vector that makes a \(45^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j}\). What angle does it make with \(\mathbf{k} ?\) c. Find a vector that makes a \(60^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? d. Is there a vector that makes a \(30^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j} ?\) Explain. e. Find a vector \(\mathbf{v}\) such that \(\alpha=\beta=\gamma .\) What is the angle?
Prove the following vector properties using components. Then make a sketch to illustrate the property geometrically. Suppose \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in the \(x y\) -plane and a and \(c\) are scalars. $$(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})$$
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