|
|
|
STEP III 1997 question 5 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1997 question 5 solution Wlog, the wheel has radius 1. Let the angle turned through be t. If the wheel wasn't rolling, the position of a point on the rim is (-sin t, 1-cos t). As the wheel is rolling, we have to add t to the x coord to get (t-sin t, 1 - cos t). Distance travelled by point on rim = Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1209x60
\displaystyle \int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt = \int_0^{2\pi}\sqrt{(1-\cos t)^2+\sin^2 t} dt \\ = \int_0^{2\pi}\sqrt{2 - 2 \cos t} dt = 2 \int_0^{2\pi}\sqrt{\frac{1}{2}(1-\cos t)} dt \\ (note that we must take the positive root). This = Distance traveled by center = 2 pi. So ratio is Solution by DFranklin |
|
|||||||||||||||
![\displaystyle 2\int_0^{2\pi} \sin \frac{t}{2} dt = -4\left[ \cos \frac{t}{2}\right]_0^{2\pi} = 8.](http://www.thestudentroom.co.uk/latexrender/pictures/b9/b9108204e847ac77735081f19dd043c4.png "\displaystyle 2\int_0^{2\pi} \sin \frac{t}{2} dt = -4\left[ \cos \frac{t}{2}\right]_0^{2\pi} = 8.")
