M408M Final Exam Solutions, December 14, 2006
1. Conic sections and polar coordinates. Consider the plane curve
r=10
1 + sin(θ).
a) Is this a circle, ellipse, parabola or hyperbola?
It is a parabola. If you don’t see this from the polar form of the equation,
see part (c).
b) Find the (Cartesian) coordinates of the point closest to the origin.
“Closest the the origin” means the smallest value of r, which means the
biggest denominator, which means θ=π/2. In polar coordinates our point
is (5, π/2), while in Cartesian coordinates the point is (0,5).
c) Express the equation of the curve in Cartesian coordinates.
Multiplying both sides by 1 + sin(θ) gives 10 = r+rsin(θ) = r+y, so
r= 10 −y, so r2= 100 −20y+y2. But r2=x2+y2, so x2= 100 −20y, or
y= 5 −(x2/20).
2. Parametrized curves. Consider the parametrized plane curve
x(t) = 2et,y(t) = t−e2t/2.
a) Find the arclenth of this curve between t=−2 and t= 2.
Since dx/dt = 2etand dy/dt = (a−e2t), we have (dx/dt)2+ (dy/dt)2=
4e2t+ 1 −2e2t+e4t= (1 + e2t)2, so our arclength is R2
−21 + e2tdt = 4 + (e4−
e−4)/2.
b) Find the slope of the tangent line to the curve at t= 0.
dy/dx = (dy/dt)/(dx/dt) = (1 −e2t)/2et= 0/2 = 0. The curve is
horizontal at this point.
3. Lines and planes in 3D space.
a) Find the equation (in standard form) of the line through the points (1,1,0)
and (3,0,1).
The vector along the line is (3,0,1) −(1,1,0) =<2,−1,1>, so our
line is x−3
2=−y=z−1, or equivalently x−1
2= 1 −y=zor equivalently
< x, y , z >=<1,1,0>+<2,−1,1> t. Any of these answers were worth
full credit.
b) Find the equation of the plane containing this line and the point (2,2,1).
First we need to find the vector perpendicular to the plane. This is the
cross product of <2,−1,1>and (2,2,1) −(1,1,0) =<1,1,1>. This is
<−2,−1,3>, so our plane is −2(x−1) −(y−1) + 3z= 0, or equivalently
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