okay 15 is kidna wacky, but apparently i did not do anything wrong the last time i got stuck

labs-set-2/template.tex

@@ -119,13 +119,48 @@
 }
 
 \qs{}{
+
   \begin{align*}
-    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 36\sin^2(2t)\cos(2t), -36\cos^2(2t)\sin(2t) \rangle \\
-    \| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \| \\
-    &= \sqrt{36^2 \sin^4(2t) \cos^2(2t)+36^2\cos^4(2t)\sin^2(2t)} \\ 
-    &=\sqrt{(36^2 \sin^2(2t) \cos^2(2t))\cdot (\sin^2(2t)+cos^2(2t))} \\
-    &=36\sin(2t)\cos(2t) \\
-    &=18\sin(4t) 
+    \vec{r}(t) &= 6\sin^3(2t)\,\hat{i} + 6\cos^3(2t)\,\hat{j} \\[6pt]
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}
+      &= \langle 36\sin^2(2t)\cos(2t),\; -36\cos^2(2t)\sin(2t) \rangle \\[6pt]
+      &= 36\sin(2t)\cos(2t)\,(\sin(2t)\,\hat{i} - \cos(2t)\,\hat{j})
+  \end{align*}
+
+  \begin{align*}
+    \left\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right\|
+      &= \sqrt{(36\sin^2(2t)\cos(2t))^2 + (-36\cos^2(2t)\sin(2t))^2} \\[6pt]
+      &= 36\sqrt{\sin^4(2t)\cos^2(2t) + \cos^4(2t)\sin^2(2t)} \\[6pt]
+      &= 36\sqrt{\sin^2(2t)\cos^2(2t)} \\[6pt]
+      &= 36\,|\sin(2t)\cos(2t)| \\[6pt]
+      &= 18\,|\sin(4t)|
+  \end{align*}
+
+  \begin{align*}
+    \hat{T}(t)
+      &= \frac{\dfrac{d\vec r}{dt}}{\left\|\dfrac{d\vec r}{dt}\right\|}
+       = \frac{36\sin(2t)\cos(2t)(\sin(2t)\,\hat{i} - \cos(2t)\,\hat{j})}{36|\sin(2t)\cos(2t)|} \\[6pt]
+      &= \sin(2t)\,\hat{i} - \cos(2t)\,\hat{j}
+  \end{align*}
+
+  \text{This corresponds with answer choice B.}
+}
+
+\qs{}{
+  Let $\vec{C}$ encode the components of the constant of integration such that $\vec{C}\in\mathbb{R}^3$  
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle t^4+5t^2,4t \rangle \\
+    \int\mathrm{d}\vec{r}= \int \langle t^4+5t^2,4t \rangle \mathrm{d}t \\
+    \vec{r}(t)&=\langle \tfrac{1}{5}t^5+\tfrac{5}{3}t^3,2t^2 \rangle + \vec{C}\\
+    \vec{C}&=\vec{r}(0)=\langle 5,-6 \rangle 
+  \end{align*}
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+    \implies \int_0^3{\left \langle \frac{2}{\sqrt{1+t}},-9t^2,\frac{6}{(1+t^2)^2}\right \rangle \mathrm{d}t }\\
+    \implies \int_0^3{\left \langle 2(1+t)^{-\tfrac{1}{2}},-9t^2,\frac{6}{(1+t^2)^2}\right \rangle \mathrm{d}t } 
   \end{align*}
 }