Math 53 Quiz 28 February 2008GSI: Theo Johnson-Freydhttp://math.berkeley.edu/∼theojf/Name:Time (circle one): 12:10 - 1:00 3:10 - 4:001. (4 pts) Consider the parameterized curvesx1= ty1= t2andx2= t3y2= t2Calculate the velocity vectors ~v1= hdx1dt,dy1dti and ~v2= hdx2dt,dy2dti at the intersectionpoint t = 1, and calculate their dot product.2. (6 pts) We can do the same thing in polar coordinates. Consider two curvesr1= sin tθ1= tandr2=√3 cos tθ2= tIn polar coordinates, we can still calculate velocities. Calculate the velocity vectors~v1= hdr1dt,dθ1dti and ~v2= hdr2dt,dθ2dti at the point of intersection t = π/3.However, the dot product between velocity vectors in polar coordinates is more com-plicated, because it depends on the value of r(t) at the point of intersection:hdr1dt,dθ1dti · hdr2dt,dθ2dti =dr1dtdr2dt+ (r(t))2dθ1dtdθ2dtCalculate this dot-product for these two curves.13. (1 pt bonus) Interpret the value of the dot-product in the previous question geomet-rically: what is it telling you about the two vectors?Please use extra paper as necessary. For each part, partial credit will be assigned based oncorrect work (you do need to show some work, enough so that I know how you solved theproblem). Please simplify and box your answers.1. We differentiate — v1(t) = hdx1dt,dy1dti = h1, 2ti and v2(t) = h3t2, 2ti — and evaluateat the intersection t = 1 to get v1= h1, 2i and v2= h3, 2i . Then the dot product ofthese vectors is v1· v2= 1 · 3 + 2 · 2 = 7 .2. We differentiate — v1(t) = hdr1dt,dθ1dti = hcos(t), 1i and v2(t) = h−√3 sin(t), 1i — andevaluate at the intersection t = π/3 to get v1= h12, 1i and v2= h−√3√32, 1i = h−32, 1i .The dot-product is evaluated at t = π/3, so r(t) = r1(t) = r2(t) =√32. Thenv1· v2=12−32+√3221 · 1 = −34+34= 0 .3. The dot product of two vectors is 0 if and only if the two vectors are perpendicular.Then dot product is positive when the angle between the vectors is less than 90◦, andnegative when the angle is more than