Applications to Geometry
We will cover following topics
Curves in Space
Vector analysis can be used to visualize and solve problems based on curves in space.
For example, the angle between two curves can be found by finding the angle between the tangents at the point of intersection.
If curves f1(x) and f2(x) intercept at P(x0,y0), then,
tanφ=ft2(x0)−ft1(x0)1+f′1(x0)⋅ft2(x0)Angle between Two Surfaces
The angle between two surfaces at a point of intersection is the angle between their normal vectors at that point.
If we write the surfaces as f(x,y,z)=constant, the normal vectors are represented by ∇f.
Then, we use the definition of dot product to calculate the angle between the two normal vectors.
Curve Parametrization
Consider a curve γ(t)=(x(t), y(t), z(t)), parameterized by t.
The derivative vector γ′(t)=(x′(t),y′(t),z′(t)) is called the velocity vector.
The length of this vector is called the speed, sometimes also referred to as instanataneous speed. It is denoted by dsdt.
So,
dsdt=|γ′(t)|=√(x′(t))2+(y′(t))2+(z′(t))2and
ds=|γ′(t)|dtThe derivative vector of unit length is called the tangent vector T(t) and is obtained by dividing the derivative vector by its length.
Therefore,
T(t)=γ′(t)|γ′(t)|=(x′(t),y′(t),z′(t))|γ′(t)|The curves parameterized in a way such that the velocity vector has unit length are called as unit-speed curves, and such parameterization as unit-speed parameterization.
The unit speed parametrization, also known as parametrization by arc length s, for γ(t) can be obtained by using the substitution t=t(s) such that:
s(t)=∫tads=∫ta|γ′(t)|dtThus,
T(s)=γ′(s)and
|γ′(s)|=1Example: To parametrize the line (1+t,2+2t,3−t), we calculate s=s(t)=∫t0√1+4+1dt=√6t.
Thus, t=1√6s and therefore the corresponding unit length parametrized curve of (1+t,2+2t,3−t) is given by (1+1√6s,2+2√6s,3−1√6s)
Curvature
If the curve is paramaterized by arc length s, the length of the acceleration vector is called the curvature, denoted by κ(s).
Thus,
κ(s)=|T′(s)|=|γ′′(s)|Here, the acceleration vector is the derivative of velocity vector and is given by:
γ′′(t)=(x′′(t),y′′(t),z′′(t))Example: The curvature of the circle γ=(acost,asint) can be found by first converting the parameterization to unit-length parameterization, which is given by
γ=(acossa,asinsa)Therefore,
T=γ′=(−sinsa,cossa)which implies that
κ=|T′|=√1a2cos2sa+1a2sin2saThe curvature for non-unit-speed parameterization is given by:
κ=|γ′′×γ′||γ′|3The unit vector in the direction of T′ is called the principal normal vector and is denoted by N. Thus,
N(s)=T′(s)|T′(s)|=T′(s)κ(s)Also, the vector B=T×N is called the binormal vector.
It indicates the direction in which the curve departs from being a planar curve.
Torsion
Torsion measures the extent to which the curve departs from being a plane curve, and is calculated for unit-speed curves as: τ=−B′⋅N
For non-unit speed parameterization,
T=γ′|γ′|=B=γ′×γ′′|γ′×γ′′|and
τ=(γ′×γ′′)⋅γ′′′|γ′×γ′′|2Example: Let us calculate the curvature and torsion of the helix (acost,asint,at). We have,
γ′=(−asint,acost,a),and
γ′×γ′′=a2(sint,−cost,1).
Thus,
|γ′×γ′′|=√2a2,which gives us
T=1√2(−sint,cost,1)and
κ=√2a2(√2)3a3=12aSimilarly, we calculate τ using the above formula and it comes out to be 12a.
Serret-Frenet’s Formulae
For a curve parameterized by arc length s, the Serret-Frenet formulas, in their matrix form, are given as:
[T′N′B′]=[0κ0−κ0τ0−τ0][TNB]where
T is the unit tangent vector,
N is the unit normal vector,
B is the unit binormal vector,
τ is the torsion,
κ is the curvature, and
x′ denotes dxds
PYQs
Curves in Space
1) Find the angle between the tangent and a general point of the curve whose equations are x=3t, y=3t2, z=3t3 and the line y=z−x=0.
[2018, 10M]
2) Prove that the vectors →a=3ˆi+ˆj−2ˆk, →b=−ˆi+3ˆj+4ˆk, →c=4ˆi−2ˆj−6ˆk can form the sides of a triangle. Find the lengths of the medians of the triangle.
[2016, 10M]
3) Find the angle between the surfaces x2+y2+z2−9=0 and z=x2+y2−3 at (2,−1,2).
[2015, 10M]
4) Find the value of λ and μ so that the surfaces (λx2−μyz) = (λ+2)x and 4x2y+z3=4 may interesect orthogonally at (1,−1,2)
[2015, 12M]
5) Show the curve →x(t)=tˆi+(1+tt)ˆj+(1−t2t)ˆk lies in a plane.
[2013, 10M]
6) A curve in space is defined by the vector equation →r=t2ˆi+2tˆȷ−t3ˆk. Determine the angle between the tangents to this curve at the points t=+1 and t=−1.
[2013, 10M]
7) Find the constants a and b so that the surface ax2−byz=(a+2)x will be orthogonal to the surface 4x2y+z3=4 at the point (1,−1,2).
[2008, 12M]
8) Show that the volume of the tetrahedron ABCD is 16(→AB×→AC)⋅→AD. Hence find the volume of the tetrahedron with vertices (2,2,2), (2,0,0), (0,2,0) and (0,0,2).
[2005, 12M]
9) The parametric equation of a circular helix is r=acosuˆi+cuˆk, where c is a constant and u is a parameter. Find the unit tangent vector ˆt at the point u and the arc length measured form u=0. Also find dˆtds where s is the arc length.
[2005, 15M]
10) Let the position vector of a particle moving on a plane curve be r(t), where t is the time. Find the components of its acceleration along the radial and transverse directions.
[2003, 15M]
11) Find the length of the arc of the twisted curve r=(3t,3t2,2t3) from the point t=0 to the point t=1. Find also the unit tangent, unit normal n and the unit binormal b at t=1.
[2001, 12M]
Curvature and Torsion
1) Find the radius of curvature and radius of torsion of the helix x=acosu, y=asinu, z=autanα.
[2019, 15M]
2) Find the curvature and torsion of the curve
→r=a(u−sinu)ˆi−a(1−cosu)ˆj+buˆk
[2018, 12M]
3) Find the curvature vector and its magnitude at any point ¯r=(θ) of the curve ¯r=(acosθ,asinθ,aθ). Show that the locus of the feet of the perpendicular from the origin to the tangent is a curve that completely lies on the hyperboloid x2+y2−z2=a2.
[2017, 16M]
4) For the cardioid r=a(1+cosθ), show that the square of the radius of curvature at any point (r,θ) is proportion to r. Also find the radius of curvature if θ=0, π4, π2.
[2016, 15M]
5) Find the curvature vector at any point of the curve ¯r(t)=tcostˆi+tsintˆj, 0≤t≤2π. Give its magnitude also.
[2014, 10M]
6) Find κ/τ for the curve
→r(t)=acostˆi+asintˆj+btˆk[2010, 12M]
7) Show that for the space curve x=t, y=t2, z=23t3, the curvature and torsion are same at every point.
[2008, 15M]
8) Find the curvature and torsion at any point of the curve x=acos2t, y=asin2t, z=2asint.
[2007, 12M]
9) If the unit tangent vector →t and binormal →b make angles θ and ϕ respectively with a constant unit vector →a, prove that sinθsinϕdθdϕ=−κτ.
[2006, 20M]
10) Find the curvature and the torsion of the space curve:
x=a(3u−u3)
y=3au2
z=a(3u+u3)
[2005, 15M]
11) Find the radii of curvature and torsion at a point of intersection of the surface x2−y2=c2, y=xtanh(zc).
[2003, 15M]
12) Find the curvature k for the space curve x=acosθ, y=asinθ, z=aθtanα.
[2002, 15M]
Serret-Frenet’s Formulae
1) Derive the Frenet-Serret formulae. Define the curvature and torsion for a space curve. Compute them for the space curve x=t, y=t2, z=23t3. Show that the curvature and torsion are equal for this curve.
[2012, 20M]
2) Show that the Frenet-Serret formulae can be written in the form d¯Tds=¯ωׯT, d¯Nds=¯ωׯN and d¯Bdx=¯ωׯB, where ¯ω=τ¯T+k¯B.
[2004, 12M]