Practice Problems for Exam 1: Solutions
Contents
1(a)
We obtain the parametric equations x=t+1, y=1, z=2-t. We need two equations without t:
Equation 1: Solving the first and third equation for t and setting them equal gives x-1 = 2 - z
Equation 2: y=1
P = sym([1,1,2]); Q = sym([2,1,1]); R = sym([1,2,1]); syms t real L = Q-P % direction vector r = P + t*L % parametric equation of line
L = [ 1, 0, -1] r = [ t + 1, 1, 2 - t]
1(b)
d = R-P % vector d from P to R dpar = dot(d,L)/dot(L,L)*L % project vector d onto vector L S = R + d
d = [ 0, 1, -1] dpar = [ 1/2, 0, -1/2] S = [ 1, 3, 0]
1(c)
The area of the triangle is half the area of the parallelogram. We can find the area of the parallelogram either using the formula with the cross product, or the formula with the dot products.
a = Q-P b = R-P area1 = norm(cross(a,b))/2 % formula with cross product area2 = sqrt(dot(a,a)*dot(b,b)-dot(a,b)^2)/2 % formula with dot products
a = [ 1, 0, -1] b = [ 0, 1, -1] area1 = 3^(1/2)/2 area2 = 3^(1/2)/2
1(d)
The volume of the tetrahedron is 1/6 of the volume of the parallepiped. The volume of the parallepiped is the absolute value of the determinant, or the absolute value of dot(cross(a,b),c).
volume1 = det([P;Q;R])/6 volume2 = dot(cross(P,Q),R)/6
volume1 = 2/3 volume2 = 2/3
1(e)
The normal vector N is orthogonal on a=Q-P and b=R-P. Then the equation for r=[x,y,z] is dot(r-P,N)=0 or dot(r,N)=dot(P,N).
Hence we have Ax+By+Cz=D with A=N_1, B=N_2, C=N_3 and D=dot(P,N), so the equation is 1*x+1*y+1*z=4.
N = cross(a,b) D = dot(P,N)
N = [ 1, 1, 1] D = 4
1(f)
For the equation of a plane we need (i) a point on the plane and (ii) a vector N' orthogonal on the plane. The point P is on the plane. The vector N' has to be orthogonal on a=Q-P and on the vector N, hence we can obtain N' by taking the cross product of a and N.
Hence we have A'x+B'y+C'z=D' with A'=N'_1, B'=N'_2, C'=N'_3 and D'=dot(P,N'), so the equation is 1*x+(-2)*y+1*z=1.
Np = cross(a,N) Dp = dot(P,Np)
Np = [ 1, -2, 1] Dp = 1
2(a)
syms t real r = [t^3/3,2*t,t^2] v = diff(r,t) a = diff(v,t) v0 = subs(v,t,-1) % substitute t=-1 a0 = subs(a,t,-1)
r = [ t^3/3, 2*t, t^2] v = [ t^2, 2, 2*t] a = [ 2*t, 0, 2] v0 = [ 1, 2, -2] a0 = [ -2, 0, 2]
2(b)
V' is given by a_T=-2. The curvature kappa is given by a_N/V^2
a0par = dot(a0,v0)/dot(v0,v0)*v0 a0orth = a0 - a0par aT = dot(a0,v0)/norm(v0) aN = norm(a0orth) kappa = aN/dot(v0,v0)
a0par = [ -2/3, -4/3, 4/3] a0orth = [ -4/3, 4/3, 2/3] aT = -2 aN = 2 kappa = 2/9
2(c)
V = simplify(norm(v))
L = int(V,t,-1,1) % integrate V(t) for t=-1 to 1
V = t^2 + 2 L = 14/3
3
We first have to find for a(t) an antiderivative v(t) with v(0)=v0. Then we have to find for v(t) an antiderivative r(t) with r(0)=r0.
syms t real a = [1,t,-1]; r0 = [1,0,0]; v0 = [1,0,1] v = v0 + int(a,t,0,t) % integrate a from 0 to t r = expand( r0 + int(v,t,0,t) ) % integrate v from 0 to t
v0 =
1 0 1
v =
[ t + 1, t^2/2, 1 - t]
r =
[ t^2/2 + t + 1, t^3/6, - t^2/2 + t]