The forward kinematics $($FK$)$ of a robot tell you where the gripper$/$hand is a given set of joint angles. For a $2-$link robot $($illustrated above, but like your shoulder and elbow$)$ the FK are given by two equations. Let $x,y$ be the position of the hand in Cartesian coordinates with respect to the shoulder. I.e$.$ the shoulder is the origin of the coordinate frame. Let ${}_1,{}_2$ be the joint angles. Let the lengths of the two limbs be $l_1,l_2.$ The FK equations are

$x=l_{1}cos{}_1+l_{2}cos({}_1+{}_2)$

$y=l_{1}sin{}_1+l_{2}sin({}_1+{}_2)$

If the joint motors are turning, then ${}_1$ and ${}_2$ are time varying. The limbs move, and the hand moves. The forward velocity kinematics $($FVK$)$ of a robot tell you what the speed of the gripper$/$hand will be for a given set of joint angular velocities. If we differentiate the equations above with respect to time, we get the FVK equations

$x^{}=-l_{1}sin{}_1{}_1^{}-l_2({}_1^{}+{}_2^{})sin({}_1+{}_2)$

$y^{}=l_{1}cos{}_1{}_1^{}+l_2({}_1^{}+{}_2^{})cos({}_1+{}_2)$

$(7$pts$)$ Let $l_1=l_2=1^1.$ Let ${}_1=\frac{}{3},{}_2=\frac{}{2}.$ We want to solve for the inverse velocity kinematics that tell us what joint velocity we need to get a desired hand velocity $x^{}=0\mathrm{.}5,y^{}=0\mathrm{.}5.$

a$.$ Write the FVK equations as a matrix equation $x^{}=A{}^{}$

b$.$ Use Gaussian elimination to solve for ${}_1^{},{}_2^{}$ to get the desired $x^{},y^{}.$

${?}^1$ It is common for limbs to be equal lengths. Check out the two limbs of your own arm by bending your elbow so

your hand touches your shoulder. Your upper arm and lower arm limbs are the same length!