SOLUTION: Suppose we have a graph of parametric equations x=bsin(t), y=b(cos(t))^3 on where t is on [0,2pi]. A. Find the point(s) on the given graph where the slope of the tangent line is

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Suppose we have a graph of parametric equations x=bsin(t), y=b(cos(t))^3 on where t is on [0,2pi]. A. Find the point(s) on the given graph where the slope of the tangent line is       Log On


   

Question 1039917: Suppose we have a graph of parametric equations x=bsin(t), y=b(cos(t))^3 on where t is on [0,2pi].
A. Find the point(s) on the given graph where the slope of the tangent line is 3/4. (Find t values not x,y values)
B. Determine the area, in terms of b, of the region enclosed by the graph.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
A. Find the point(s) on the given graph where the slope of the tangent line is 3/4. (Find t values not x,y values).
x=bsint and y=b%28cost%29%5E3
==> dx%2Fdt+=+b%2Acost and dy%2Fdt+=+-3b%28cost%29%5E2%2Asint after differentiating both x and y wrt t.
==> dy%2Fdx+=+-3sint%2Acost after division.
To find the point(s) on the graph where the slope of the tangent line is 3/4, let dy%2Fdx+=+-3sint%2Acost+=+3%2F4 .
==> -sint%2Acost+=+1%2F4 ==> -2sint%2Acost+=+1%2F2 ==> sin2t+=+-1%2F2.
==> 2t+=+%287%2Api%29%2F6 or 2t+=+%2811%2Api%29%2F6.
==> t+=+%287%2Api%29%2F12 or t+=+%2811%2Api%29%2F12.
B. Determine the area, in terms of b, of the region enclosed by the graph.
To determine the area, first express y in terms of x.
x=bsint, y=b%28cost%29%5E3+
==> x%2Fb=sint, %28y%2Fb%29%5E%281%2F3%29=cost+
==> x%5E2%2Fb%5E2+%2B+y%5E%282%2F3%29%2Fb%5E%282%2F3%29+=+1
==> y%5E%282%2F3%29%2Fb%5E%282%2F3%29+=+1+-x%5E2%2Fb%5E2++=+%28b%5E2+-+x%5E2%29%2Fb%5E2
==> y%5E%282%2F3%29=%28b%5E%282%2F3%29%2Fb%5E2%29%2A%28b%5E2+-+x%5E2%29
==> y%5E%282%2F3%29=%281%2Fb%5E%284%2F3%29%29%2A%28b%5E2+-+x%5E2%29
==> y=%281%2Fb%5E2%29%2A%28b%5E2+-+x%5E2%29%5E%283%2F2%29 or y=%28-1%2Fb%5E2%29%2A%28b%5E2+-+x%5E2%29%5E%283%2F2%29.
Now for purpose of getting the area assume that b > 0. (The sense of the graph in this case is clockwise. b < 0 would only change the direction to ccw but would yield the same graph and the same area.)
==> Area = %284%2Fb%5E2%29int%28%28b%5E2+-+x%5E2%29%5E%283%2F2%29%2C+dx%2C+0%2Cb%29
(I wrote it this way rather than %282%2Fb%5E2%29int%28%28b%5E2+-+x%5E2%29%5E%283%2F2%29%2C+dx%2C+-b%2Cb%29. They give the same answer.)
Let x = bsint.
==> dx = bcostdt
==> Area =
=
Using %28cost%29%5E2+=+%281%2Bcos2t%29%2F2,
Area = b%5E2%2Aint%28%281%2B2cos2t%2B%28cos2t%29%5E2%29%2C+dt%2C+0%2Cpi%2F2%29+
=b%5E2%2Aint%28%281%2B2cos2t%2B%281%2Bcos4t%29%2F2%29%2C+dt%2C+0%2Cpi%2F2%29+
=b%5E2%2Aint%28%283%2F2%2B2cos2t%2B%281%2F2%29%2Acos4t%29%2C+dt%2C+0%2Cpi%2F2%29+
=b%5E2%2A%28%283%2F2%29t%2B+sin2t+%2B+%281%2F8%29%2Asin4t%29%5B0%5D%5E%28pi%2F2%29
=b%5E2%2A%283%2F2%29%2A%28pi%2F2%29+=+%283b%5E2%2Api%29%2F4