Chapter 11: Problem 3
Describe the plane \(x=4\)
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Chapter 11: Problem 3
Describe the plane \(x=4\)
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Compute the following derivatives. $$\frac{d}{d t}\left(t^{2}(\mathbf{i}+2 \mathbf{j}-2 t \mathbf{k}) \cdot\left(e^{t} \mathbf{i}+2 e^{t} \mathbf{j}-3 e^{-t} \mathbf{k}\right)\right)$$
Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Algebra inequality Show that $$\left(u_{1}+u_{2}+u_{3}\right)^{2} \leq 3\left(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\right)$$ for any real numbers \(u_{1}, u_{2},\) and \(u_{3} .\) (Hint: Use the CauchySchwarz Inequality in three dimensions with \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and choose v in the right way.)
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. $$\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$$
Prove that \(|c \mathbf{v}|=|c||\mathbf{v}|,\) where \(c\) is a scalar and \(\mathbf{v}\) is a vector.
Properties of dot products Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle,\) and \(\mathbf{w}=\left\langle w_{1}, w_{2}, w_{3}\right\rangle .\) Prove the following vector properties, where \(c\) is a scalar. $$|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u} \| \mathbf{v}|$$
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