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I figured out the issue. You took the question to mean the target is $14$ metres horizontally and $5.5$ metres vertically from the exact point of release. The textbook answer is based on the target being $14$ metres horizontally and $5.5$ metres vertically from the point on the ground directly (vertically) below the point of release. In other words, $14$ ...


3

Velocity is usually a vector, and speed a scalar (the modulus of the velocity). So if you are picky on the terms, only the book is right.


2

$\def\o{{\large\tt1}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$For typing convenience, define $\o\in{\mathbb R}^{n}$ as the all-ones vector and the matrices $$\eqalign{ M &= FZ^TZ \quad&\implies\quad M_{ij} = F_{ip}Z_{qp}Z_{qj} \\ \beta& \quad&\implies\quad \beta_{ij} = \o_k\,\alpha_{ikj} \\ }$$ NB: $\,$ For the equations to be ...


2

It may help to use the index notation, $$ f(\vec x) = \exp(-x_i t_i) $$ Then $$ \frac{d f}{dx_j} = -\exp(-x_i t_i) t_i \delta_{ij} = -\exp(-x_i t_i)t_j, $$ where $\delta_{ij} = 1$ if $i = j$ and zero otherwise. This can be written compactly as $\nabla f = -\exp(-\langle x,t \rangle) t$. Taking a second derivative $$ \frac{d^2 f}{dx_j dx_k} = \exp(-x_i t_i)...


2

$\vec F=(e^{yz}, 0, 0)$ So as you found, $\displaystyle \nabla \times \vec F = (0, ye^{yz}, -ze^{yz})$ Now the surface is $x^2+y^2+z^2 = 1, z \geq 0$. So the unit normal vector to the surface is $\hat n = (x, y, z)$. So the surface integral is $\displaystyle \iint_\Sigma (\nabla \times \vec F) \cdot \hat n \ dS = \iint_\Sigma (y^2-z^2) e^{yz} \ dS$ which ...


1

You have a curve $X(t)$ parametrized by the variable $t$ in the interval $I = [a,b]$, but you want to reparametrize the curve, using a new parameter $s$, which lives in a new interval $J$, so that if changing $s$ by 1 unit, moves along the curve by 1 distance unit (measured by the arc length formula). This theorem is maybe known as the "arc-length ...


1

You didn't talked about the domain of the functions. Lets say, for the sake of simplicity, that $f:[a,b]\times [c,d] \to \mathbb R$. Then the graph of $f(x,y)$ is the set $\{(x,y,f(x,y)) \in \mathbb R^3\mid x\in [a,b], y \in [c,d]\}$. If the function is continuous and positive, its graph would be a surface above the rectangle $[a,b]\times [c,d]$ If you ...


1

The two expressions mean the same thing. The latter just specifies what the variables $x,y$ depend on. In this case time. You can think of as $x$ and $y$ vary in time, they draw out the surface $f(x,y)=f(x(t),y(t))$ as $t$ varies. I hope this helps with your understanding :).


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HINT Let us assume the inner product is defined over a complex vector space. Then we have: \begin{align*} \begin{cases} \langle a + b,a + b\rangle = \langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle + \langle b,b\rangle\\\\ \langle a - b,a - b\rangle = \langle a,a\rangle - \langle a,b\rangle - \langle b,a\rangle + \langle b,b\rangle \end{cases} \...


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