Use a graph to estimate the y-coordinate of the highest points on the given curve. Then use calculus to find the exact value.?

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 ..... r

0 ..... 0

0.05..... .099

0.1 ..... 0.198

0.015..... 0..295

and 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 .

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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 .....

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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 = 0

so, numerator only :

cos(2θ) .2 sin(θ) + sin(2θ) cos(θ)] = 0

Now 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) = 0

or

sin(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) ] = 0

or sin(θ) [ 3cos(2θ) + 1 ] = 0

By the zero product property :

either sin(θ) = 0 which occurs at θ = 0 , (which is a minimum for r )

or 3cos(2θ) +1 = 0

so 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...)