Tricky calc 2 question. How do you cut a pizza into 3 slices of equal area with 2 parallel cuts? Radius is 7. I have gotten this far: I took the integral from 0 to c, of the equation of a circle y=sqrt(49-x^2) and ultimately set it equal to the area of 1/3rd of the Quadrant I quarter. I solved the integral using trig substitution but end up with an equation that I can't solve for c, without a calculator. WE are supposed to get to an approx. solution without a calculator/graphing calculator. She said if we get to the point I am at, we have went the wrong way. I'm baffled. Anyone?
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\[1/2(c \sqrt{49-c^2}+49\sin^{-1} (\frac{ c }{ 7 }))=49\pi/12\]the integral solved is:
this came from \[\int\limits_{0}^{c}\sqrt{49-x^2}dx\] using trig sub x=7sin(theta)
I have an idea, you consider whether that helps or not. I consider just 1/4 of the circle, at quadrant 1 and take integral of the circle from 0 to 7. by that way, i get the area under the curve and divided by 3 to get 1/3 part of that .from that find out the "cut" to improve it into other parts
I actually thought of this, I hadn't executed it tho. The issue is, when I evaluate it, will it give me the point where the cut needs to be made? If not, how do I get the point from what I solve?
A = pi * r^2 = 49 pi make integral from c to 7 = 49 pi and solve for c
sorry , divide by 3
what? how can I solve it with a c in it? I can't use a calculator
do you try go backward like this: after take integral, you have something and replace 7 and 0 to that to get the answer, now you have 1/3 of Area, get c as POstrOn says you solve for c. I agree with him
hmm, oky, lemme attempt that
find the integral and split it with F(7) - F(c)
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