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10.4 - Reading questions
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<h2>[10.4] - Reading Assignment
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<p>After reading section 10.4...


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<li>
What is the relationship between the velocity vector and the tangent vector?
<p class="justme answer">The velocity vector $\myv r'(t)$ is always tangent to the space curve ("trajectory") $\myv r(t)$.  So $\myv r'$ <b>is parallel</b> to any tangent vector.  It is a scalar multiple of any 
<li>A particle moves along the curve defined by the equation $y = \sin (\pi x)$. The $x$-coordinate, $ x (t)$ of the
particle satisfies the equation $\frac{dx}{dt}=e^{2t}$. At $t=0$ the curve is at the point $(\frac12,1)$.
<ol type="a">
<li>Find $x(t)$ in terms of $t$.
<p class="justme answer">$$\begineq x(t)-x(0)&=&\int_0^t e^{2t'}\,dt'=\left.\frac 12e^{2t'}\right|^{t'=t}_0\\
x(t)-\frac 12&=&\frac 12\left(e^{2t}-e^0\right)=\frac 12e^{2t}-\frac 12\\
x(t)&=&\frac 12e^{2t}\endeq$$


<li>Find $y(t)$ in terms of $t$.
<p class="justme answer">$$y(x(t))=\sin(\pi x(t))=\sin(\frac12\pi e^{2t}).$$

<li>Find the velocity vector, $\myv v(t)$.
<p class="justme answer">$$\myv v(t)=\myv r'(t)=\langle x'(t), y'(t) \rangle$$
And we already have $\frac{dx}{dt}=e^{2t}$. And 
$$y'(t)=\frac{d}{dt}\sin\left(\frac12\pi e^{2t}\right)= \cos\left(\frac12\pi e^{2t}\right) \, \pi e^{2t}$$
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<li>Muddy questions? Questions you wonder about?
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