diff --git a/mr/ubB/mrB.pdf b/mr/ubB/mrB.pdf
index 3d26584..090a350 100644
--- a/mr/ubB/mrB.pdf
+++ b/mr/ubB/mrB.pdf
Binary files differ
diff --git a/mr/ubB/mrB.tex b/mr/ubB/mrB.tex
index 3e9d0ce..481d02b 100644
--- a/mr/ubB/mrB.tex
+++ b/mr/ubB/mrB.tex
@@ -88,7 +88,7 @@
             \item $(s)\rightarrow s:(b,e)\rightarrow b:(e,a,c)\rightarrow e:(a,c,f,g)\rightarrow a:(c,f,g)\\\rightarrow c:(f,g,d)\rightarrow f:(g,d,h,z)\rightarrow z$ found $path=(s,e,f,z)$
             \item $((s,0))\rightarrow s: ((b,1),(e,1))\rightarrow b: ((e,1),(a,2),(c,7))\rightarrow e: ((a,2),(g,2),(c,7),(f,7)) \\\rightarrow a:((g,2),(c,5),(f,7))\rightarrow g: ((h,3),(c,5),(f,7),(z,8))\\\rightarrow h:((z,4),(c,5),(f,7))\rightarrow (z,4), path=(s,e,g,h,z)$
             \item $(s,0+2)\rightarrow s:((e,1+1),(b,1+3))\rightarrow e:((g,2+0),(b,1+3),(f,7+1))\\\rightarrow g:((b,1+3),(h,3+1),(z,8+0))\rightarrow b:((h,3+1),(a,2+4),(z,8+0),(c,7+3))\\\rightarrow h:((z,4+0),(a,2+4),(c,7+3))\rightarrow z: path=(s,e,g,h,z)$\\
-           
+
         \end{enumerate}
     \Aufgabe{Nonholonomic Constraints}{4}
         \begin{enumerate}[(a)]
@@ -102,22 +102,20 @@
                 0&0&1
             \end{pmatrix}\dot\vec q\\
             =\begin{pmatrix}
-                \sin\theta & \cos\theta & 0
+                -\sin\theta & \cos\theta & 0
             \end{pmatrix}\dot\vec q \\
-            = \dot x \sin\theta + \dot y \cos\theta\\=0$
+            = -\dot x \sin\theta + \dot y \cos\theta\\=0$
 
             \item nonholonomic, because the derivative of q is used\\
         \end{enumerate}
     \Aufgabe{Reeds-Shepp-Curves}{8}
         \begin{enumerate}[(a)]
             \item As the tangent is always perpendicular to the radius there are right triangles. So for $r_1=r_2=r$ we can calculate:\\
-
             $d=2\cdot \sqrt{\left(\frac{1}{2}dist(\mathbf{s'},\mathbf{g'})\right)^2-r^2}$\\
             For $r_1\not=r_2$ we can use:\\
             $d=\sqrt{dist(\mathbf{s'},\mathbf{g'})^2-(r_1+r_2)^2}$\\
             There geometrical explanation is to translate $r_2$ and append it to $r_1$. Then there is only one right triangle and we again can calculate the pythagorean.
             \item $\delta(\alpha_r,\mathbf{s'},\mathbf{g'},\alpha_l)=r\alpha_r+d+r\alpha_l=r\alpha_r+2\cdot\sqrt{\left(\frac{1}{2}dist(\mathbf{s'},\mathbf{g'})\right)^2-r^2}+r\alpha_l$
-
             \item $\delta(\alpha_r,\mathbf{s'},\mathbf{g'},\alpha_l)=r_1\alpha_r+d+r_2\alpha_l=r_1\alpha_r+\sqrt{dist(\mathbf{s'},\mathbf{g'})^2-(r_1+r_2)^2}+r_2\alpha_l$
             %passt
         \end{enumerate}