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# System prepended metadata

title: HW7

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1.

2. Let $\vec a , \vec b \in \mathbb{R}^3$.
    a. Show that for all vectors $\vec a$ and $\vec b$, $$4(\vec a \cdot \vec b) = \|{\vec a + \vec b}\|^2 - \|{\vec a - \vec b}\|^2.$$
    >(sol) Note that $$\begin{aligned}\|\vec{a}+\vec{b}\|^2 &= (\vec{a} +\vec{b})\cdot (\vec{a} + \vec{b}) = \|\vec{a}\|^2 + 2\vec{a} \cdot \vec{b} + \|\vec{b}\|^2 \\ \|\vec{a}-\vec{b}\|^2 &= (\vec{a} -\vec{b})\cdot (\vec{a} - \vec{b}) = \|\vec{a}\|^2 - 2\vec{a} \cdot \vec{b} + \|\vec{b}\|^2\end{aligned}$$ Substract the second equation from the first equation, we obtain the equation in the question.

    b. Show that $\vec a \perp \vec b$ iff $\|{\vec a + \vec b}\| = \|{\vec a - \vec b}\|$.
    > (sol) By question a., $$\vec{a}\perp\vec{b} \iff \vec{a}\cdot \vec{b} = 0\iff \|{\vec a + \vec b}\|^2 - \|{\vec a - \vec b}\|^2 = 0 \iff \|{\vec a + \vec b}\| = \|{\vec a - \vec b}\|$$
 
    c. Show that, if $\vec a$ and $\vec b$ are nonzero vectors such that $$(\vec a + \vec b) \perp (\vec a - \vec b) \qquad \text{and} \qquad \|{\vec a + \vec b}\| = \|{\vec a - \vec b}\|,$$ then the parallelogram generated by $\vec a$ and $\vec b$ is a square.
    > (sol) By the first condition, $$0 =(\vec a + \vec b)\cdot (\vec a - \vec b) = \|\vec a\|^2 - \|\vec b\|^2$$Hence $\|\vec a\| = \|\vec b\|$.
    > Follows the question b., the second condition equivalent to $\vec a \perp \vec b$. Combine these two result, the parallelogram generated by $\vec a$ and $\vec b$ is a square.

3. Show that $$|\vec a \cdot \vec b| \leq \|{\vec a}\| \|{\vec b}\|,$$ and the equality holds iff $\vec a$ and $\vec b$ are **parallel**, i.e. there exists $\lambda$ such that $\vec a = \lambda \vec b$ or $\vec b = \lambda \vec a$.
    >(sol 1) The case $\vec b = 0$ or $\vec a = 0$ is trivial. For the other case, we know that $\vec a \cdot \vec b = \|\vec a\|\|\vec b\|\cos \theta$. The inequality follows by $\cos \theta \leq 1$. Moreover, the equality holds for $\cos \theta = \pm 1$, hence $\vec a$ and $\vec b$ are parallel, that is, $\exists \ \lambda$ such that $\vec a = \lambda \vec b$. 

    >(sol 2) The case $\vec b  = 0$ or $\vec a = 0$ is trivial. For the other case, consider vector $$\vec u = {\vec a \cdot \vec b\over \|\vec b\|^2} \, \vec b$$ We have that $$0 \leq \|\vec a - \vec u\|^2 = \|\vec a\|^2 - 2 \vec a \cdot \vec u + \| \vec u\|^2$$ Note that $\|\vec u\|^2 = \vec a \cdot \vec u = {|\vec a \cdot \vec b|^2\over \|\vec b\|^2}$. Hence $${|\vec a \cdot \vec b|^2 \over \|\vec b \|^2}\leq \|\vec a\| \quad (\implies |\vec a \cdot \vec b| \leq \|\vec a\| \|\vec b \|)$$The equality holds if and only if $\vec a = \vec u = \lambda \vec b$ where $\lambda = {\vec a \cdot \vec b\over \|\vec b \|^2}$.

4. Prove the **parallelogram law**: $$\|{\vec a + \vec b}\|^2 + \|{\vec a - \vec b}\|^2 = 2 \|{\vec a}\|^2 + 2 \|{\vec b}\|^2.$$
    >(sol) Add up two equation in solution of question 2a.

5. 

6. 

7. Let $\vec f$ be a differentiable vector-valued function. Show that if $\|\vec{f}(t)\| \neq 0$, then $${d\over dt}(\|\vec{f}(t)\|) = {\vec{f}(t)\cdot\vec{f}'(t)\over \|\vec{f}(t)\|}$$
    > (sol) Compare $${d\over dt}(\|\vec{f}(t)\|^2) = 2\|\vec{f}(t)\|{d\over dt}(\|\vec{f}(t)\|)$$and $${d\over dt}(\vec{f}\cdot \vec{f}) = 2\vec{f}(t)\cdot \vec{f}'(t)$$

8. Show that $\|\vec{\gamma}(t)\|$ is constant if and only if $\vec{\gamma}(t)\cdot \vec{\gamma}'(t) = 0$ for all $t$.
    > (sol) Differentiate $\vec{\gamma}\cdot \vec{\gamma}$.

9. 

10. Let $$\vec\gamma(t) = 3\cos t \vec{i} + 3 \sin t\vec j + 4t \vec k, \qquad t \geq 0.$$
 
    a. Let $$ s(t) = \int_0^t \|{\vec\gamma'(u)}\| \, du.$$ Show that $s$ is a one-to-one function, and find its inverse function $\tau(s)$.
    >(sol) By direct computation, we obtain $\|\vec{\gamma}'(t)\| = 5$. So $s(t) = 5t$ which is obviously one-to-one. And the inverse function is $\tau(s) = {1\over 5}s$ for $s\geq 0$.
    >==Remark== Note that the function $s$ is always increasing.

    b. Let $$\vec R(s) = \vec\gamma(\tau(s)).$$ Show that $$\left\|{\dfrac{d \vec R}{ds}}\right\| =1.$$ (The parametrization $s$ is called the **parametrization by arc length**.)
    >(sol) Using chain rule $${d\vec{R}\over ds} = {d\vec{R}\over d\tau}{d\tau \over ds} = {1\over 5}(-3\cos \tau \vec{i} + 3\cos \tau \vec{k} + 4\vec k)$$Hence the norm is $${1\over 5}\sqrt{(3^2\cos^2\tau + 3^2 \sin^2\tau + 4^2)} = 1$$

    c. Find the length of the curve $$\vec R(s), \qquad 0 \leq s \leq L.$$
    >(sol) By the formula of arc length $$\int_{0}^{L}\|{d\vec{R}\over ds}\| \ ds = \int_{0}^{L}1 \ ds = L$$
    > ==Remark== You see why this parametrization is "by arc length". In short, with this parametrization, the arc length is the same as how long does parameter $s$ go.