Updated spacing between @ latex expressions

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revision #10352 parent #9952 GitHub c702ebe ← older newer →
en/1.3.13.md +8 −8
@@ -56,11 +56,11 @@
<h3>Solution</h3>
<p>
<p>a) We will use the formula for the flight path obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:</p>
−$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2 \varphi} $$
+$$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2\cos^2 \varphi} $$
<p>Let's use the <a href="#hint">hint</a>,</p>
−$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\quad(1) $$
+$$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2}(1+\tan^2 \varphi)\quad(1) $$
<p>Where do we express the angle $ \varphi$:</p>
−$$ \tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx} $$
+$$ \tan\varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx} $$
<p>b) Let's depict a set of trajectories, changing the angle $\varphi$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:</p>
</p>
<center>
@@ -83,12 +83,12 @@
$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2} $$
$$ x(t) = vt \cdot \cos{\varphi} $$
−<p>Let $\xi=\cos{\varphi}$, </p>
−$$ v_y = v \xi $$
+<p>Let $\xi =\cos{\varphi}$, </p>
+$$ v_y = v \xi$$
−$$ v_x = v \sqrt{1-\xi ^2} $$
+$$ v_x = v \sqrt{1-\xi^2} $$
<p>Getting a function</p>
−$$ v(\xi)=\frac{\sqrt{g}x}{\sqrt{2 \xi \sqrt{1-\xi ^{2}}x-\xi ^{2}y}} \quad(2) $$
+$$ v(\xi )=\frac{\sqrt{g}x}{\sqrt{2 \xi\sqrt{1-\xi^{2}}x-\xi^{2}y}} \quad(2) $$
<p>The maximum of the radical value corresponds to the minimum of the function (2)</p>
$$ (\sqrt{x^{2}+y^{2}-y})/2 \rightarrow \max $$
<p>Minimum of function $(2)$</p>
@@ -98,7 +98,7 @@
<h4>Answer</h4>
<p>
− $$\tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx}.$$
+ $$\tan\varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx}.$$
$$y=\frac{v^2}{2g}-\frac{gx^2}{2v^2}$$
$$v_{\text{min}}=\sqrt{g(y+\sqrt{x^2+y^2})}$$
</p>
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