You can plot a polar curve like this by hand . Its tedious , and needs a good protractor and measurement device . The best jump is probably a θ -step of about 0.5 or so.Your Data table will look something like this :θ ..... r0 ..... 00.05..... .099 0.1 ..... 0.1980.015..... 0..295and so on ..However, seeing as they are pretty common, using something like a TI-84 device , set in polar mode , and plotting :r1 = sin(2θ ) on a window of {0,6.28, 0.05, -1 ,1 ,1, -1, 1 , 1} will give a MUCH faster picture of this " 4-leaved Rose" curve .=========================Using a polar sketch , and "TRACE" you can find two symmetric maxima :at ( θ , x , y) = ( .95 , +0.557 , 0.769) and another at ( 3.3 , -.557 , 0.769)Once you have found one by calculus , you can easily find the next one .So now we know exactly what we are looking for .....======================== Solution to finding the exact value :The polar curve is r = sin (2θ )and to convert to x-y co-ordinates, build a right triangle , with base angle θ , hypotenuse of r and x and y in the usual places.The conversions are y = r sin(θ ) and x = r cos( θ ) but for this problem , r = sin(2θ)so the conversions from polar to x-y become y = sin(2θ ) sin(θ) and x = sin(2θ) cos(θ).....(1)To find the maximum on an x-y plot, the standard process is to find dy/dx and set it = 0 but we do not have y or x as explicit functions of x , only of θ so.....Differentiating top part : dy/dθ = d/dθ [sin(2θ ) sin(θ)] = cos(2θ) .2 .sin(θ) + sin(2θ) cos(θ)( using the product rule and the chain rule )and similarly for the bottom part :dx/dθ = cos(2θ) .2 .cos(θ) + sin(2θ){-sin(θ)}and now the " magic "dy/dx = [dy/dθ] / [dx /dθ]so we get this huge fraction :dy/dx = [cos(2θ) .2 .sin(θ) + sin(2θ) cos(θ)] / [cos(2θ) .2 .cos(θ) + sin(2θ){-sin(θ)}]which is to be set = 0 to find the maximum .Recall from Math 8 /9 that only the numerator has to be set = 0so, numerator only :cos(2θ) .2 sin(θ) + sin(2θ) cos(θ)] = 0Now at this point , you could use a graphing calculator which you are already using , and plot, on x-y function mode y1 = cos(2x .2 sin(x) + sin(2x) cos(x)] and find where it crosses the axis , which occurs at x = 0.95... or so But it can be solved exactly :Factoring out a " sin(θ) " and using sin(2θ) = 2sin(θ) cos(θ) the expression becomes :2cos ( 2θ) sin(θ) + 2 sin(θ) cos(θ) cos(θ) = 0orsin(θ) [2cos(2θ) + 2 ((cos(θ)) ^2 ] = 0 but there is an identity from Math 11 :2(cos(θ)) ^2 = cos(2θ) +1 so the expression to be solved is sin(θ) [ 2cos(2θ) + 1 + cos(2θ) ] = 0or sin(θ) [ 3cos(2θ) + 1 ] = 0By the zero product property :either sin(θ) = 0 which occurs at θ = 0 , (which is a minimum for r )or 3cos(2θ) +1 = 0so cos(2θ) = -(1/3) and 2θ = arc cos ( -1/3 ) = cos^(-1) (-1/3) = 1.91063...... from a calculator and ( finally ) θ = (1.91063... )/ 2 = 0.955316... <--- answer for θ at the required spot Now if you just want the " y " value at that θ : use equation (1) from way back :y = sin(2θ ) sin(θ) = y = sin(2(.955316... ) sin(.955316...)or y = 0.7698003...<---- answer for the y max.and the problem is done . And similarly , again from equation (1):x = sin(2θ) cos(θ) = sin(2.(955...) cos(.955...)and this gives x max = 0.544...Both this θ and x and y agree with the original TRACE done on the first sketch ..To find the symmetric max at the left hand leaf, just use x = - 0.5443... and the same y value : ( 3.3097 , - 0.5443... , 0.7698...)...
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