Calc3: Geometrics
Equations of Lines and Planes
Lines
Given r=(x,y,z);v⃗=<a,b,c>r = (x,y,z); \vec v =<a,b,c>r=(x,y,z);v=<a,b,c>. We can write a point rrr on a line as
r=r0+t⋅v⃗r = r_0 + t \cdot \vec v r=r0+t⋅v
Another way of writing a line. Assume a,b,c are non-zeros, then we can solve for t using
t=x−x0a=y−y0b=z−z0ct = \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} t=ax−x0=by−y0=cz−z0
The equation above is called symmetric equations.
We can rewrite as parametric equation.
x=x0+aty=y0+btz=z0+ctx = x_0 + at\\\\ y = y_0 + bt\\\\ z = z_0 + ct\\\\ x=x0+aty=y0+btz=z0+ct
To describe a line segment, imagine there is a particle. At t=0, this particle is at r0r_0r0. It moves at a constant speed so that it arrives at r1r_1r1 at t=1. The trace is the line segment. This is called vector equation.
r0+t(r1−r0)r_0 + t(r_1 - r_0) r0+t(r1−r0)
Planes
There are two options to determine a plane.
The first option:
Assume plane passes through r0r_0r0 with normal vector u⃗\vec uu. For an arbitrary point r on this plane, we know u⃗⋅r0r⃗=0\vec u \cdot \vec{r_0 r} = 0u⋅r0r=0. We can expand the equation as
d=−(ax0+by0+cz0)ax+by+cz+d=0d = -(ax_0+by_0+cz_0) \\\\ ax+by+cz+d = 0 \\\\ d=−(ax0+by0+cz0)ax+by+cz+d=0
The equation above is called linear equation.
The second option:
Assume three points P, Q, R stay on a plane. We can first calculate PQ⃗\vec{PQ}PQ and PR⃗\vec{PR}PR, and use cross product to compute a normal vector. The problem is converted to the first case.
Angle Between Planes
Two planes are parallel if they have parallel normal vectors. The normal vector of a plane ax+by+cz+d=0ax+by+cz+d=0ax+by+cz+d=0 can be represented by <a,b,c><a,b,c><a,b,c>.
If two planes are not parallel, there is an angle between these two planes.
Say the normal vector of the first plane is n⃗1\vec n_1n1, and that of the second is n⃗2\vec n_2n2. The angle between these two normal vectors is θ\thetaθ, which is supplementary to the angle of two planes α\alphaα.
cosθ=n⃗1n⃗2∣n⃗1∣∣n⃗2∣α=180∘−θcos \theta = \frac{\vec n_1 \vec n_2} {|\vec n_1| |\vec n_2|}\\\\ \alpha = 180^\circ - \theta cosθ=∣n1∣∣n2∣n1n2α=180∘−θ
To determine the intersection line, we need one point and the direction. Usually, we set z=0. Solve the equation set. Then, we calculate the cross product of those two normal vectors, which is the direction.
Distance between Point and Planes
Say we have a point P1=(x1,y1,z1)P_1=(x_1, y_1, z_1)P1=(x1,y1,z1) and we want to find its distance to a plane ax+by+cz+d=0ax+by+cz+d=0ax+by+cz+d=0. Find an arbitrary point on the plane P0=(x0,y0,z0)P_0=(x_0, y_0, z_0)P0=(x0,y0,z0) on the plane. The point AAA is the projection of P1P_1P1 on the plane.
The length of P1AP_1AP1A is the projection of P0P1P_0P_1P0P1 along the normal vector, which is the distance desired.
∣P1A∣=∣P0P1⃗⋅n⃗∣∣n⃗∣=∣(ax1+by1+cz1)−(ax0+by0+cz0)∣a2+b2+c2=∣ax1+by1+cz1+d∣a2+b2+c2|P_1A| = \frac{|\vec{P_0P_1} \cdot \vec{n}|}{|\vec{n}|} \\\\ = \frac{|(ax_1+by_1+cz_1)-(ax_0+by_0+cz_0)|}{\sqrt{a^2+b^2+c^2}} \\\\ = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \\\\ ∣P1A∣=∣n∣∣P0P1⋅n∣=a2+b2+c2∣(ax1+by1+cz1)−(ax0+by0+cz0)∣=a2+b2+c2∣ax1+by1+cz1+d∣
The distance between two planes can be found by fixing a point on one of the two planes, and then calculating the distance between that point to the other plane.
Polar Coordinate
Connections to Cartesian
Convert P=(r,θ)P=(r, \theta)P=(r,θ) to the Cartesian coordinate is (rcosθ,rsinθ)(rcos\theta, rsin\theta)(rcosθ,rsinθ).
Covert P=(x,y)P = (x,y)P=(x,y) to the Polar coordinate is (x2+y2,arctanyx)(\sqrt{x^2+y^2}, arctan\frac{y}{x})(x2+y2,arctanxy).
Graphics
ex.1
Sketch the curve r=2cosθr=2cos\thetar=2cosθ, range for θ∈[0,π/2]\theta \in [0, \pi/2]θ∈[0,π/2].
We know from the connections to cartesian
x=rcosθcosθ=x/rx = rcos\theta \\\\ cos\theta = x/r \\\\ x=rcosθcosθ=x/r
Plug into the curve function,
r2=2x(x−1)2+y2=1r^2 = 2x \\\\ (x-1)^2 + y^2 = 1 r2=2x(x−1)2+y2=1
So, it is a circle of radius 1 centered at (1,0).
Tangent Line
Interpretation in physics: direction of velocity at a given point.
In Polar coordinate, we have a curve r=f(θ)r = f(\theta)r=f(θ). Express this curve in Cartesian coordinate, and then compute the tangent.
dydx=dy/dθdx/dθ=dr/dθ⋅sinθ+r⋅cosθdr/dθ⋅cosθ−r⋅sinθ\frac {dy}{dx} = \frac {dy/d \theta}{dx/d \theta} \\\\ = \frac {dr/d \theta \cdot sin \theta + r \cdot cos \theta }{dr/d \theta \cdot cos \theta - r \cdot sin \theta} dxdy=dx/dθdy/dθ=dr/dθ⋅cosθ−r⋅sinθdr/dθ⋅sinθ+r⋅cosθ
Cylinders and Quadric Surfaces
Quadratic Forms and Reductions
The general form for quadratic terms are listed below
Ax2+By2+Cz2+Dxy+Eyz+Fxz+Gx+Hy+Iz+J=0Ax^2 + By^2+Cz^2 + Dxy+Eyz+Fxz+ Gx+Hy+Iz + J = 0 Ax2+By2+Cz2+Dxy+Eyz+Fxz+Gx+Hy+Iz+J=0
By translation and rotation, we have two standard forms
Ax2+By2+Cz2+J=0Ax2+By2+Iz=0Ax^2 + By^2+Cz^2 + J =0 \\\\ Ax^2 + By^2 + Iz = 0 Ax2+By2+Cz2+J=0Ax2+By2+Iz=0
Sketching
ex. 1
We have a surface x2+y2/9+z2/4=1x^2 + y^2/9 + z^2/4 = 1x2+y2/9+z2/4=1.
Sol:
When z=0, it is an ellipse. For a general plane z=k,
x2+y2/9=1−k2/4x^2 + y^2/9 = 1 - k^2/4 x2+y2/9=1−k2/4
Notice that the domain for k is [−2,2][-2,2][−2,2].
This is an ellipsoid.
ex. 2
We have a surface z=4x2+y2z = 4x^2+y^2z=4x2+y2.
Sol:
When z=0, we have both x and y to be 0. For a general plane z=k,
k=4x2+y2k = 4x^2+y^2 k=4x2+y2
This is an elliptic paraboloid.
ex. 3
We have a surface z=y2−x2z = y^2-x^2z=y2−x2.
Sol:
When x=k, z=y2−k2z = y^2 - k^2z=y2−k2, which is a parabola, pointing upwards. When y=k, z=k2−x2z = k^2-x^2z=k2−x2, which is a parabola, pointing downwards.
This looks like a saddle, called hyperbolic paraboloid.
ex. 4
We have a surface x2/4+y2−z2/4=1x^2/4+y^2-z^2/4 = 1x2/4+y2−z2/4=1.
This is a hyperboloid of one sheet.
ex. 5
We have a surface x2+2z2−6x−y+10=0x^2+2z^2-6x-y+10 = 0x2+2z2−6x−y+10=0. First complete squares and write the quadratic form as
(x−3)2+2z2−y+1=0(x-3)^2 + 2z^2 - y +1 = 0 (x−3)2+2z2−y+1=0
Substitute x=x−3x = x-3x=x−3.
x2+2z2−y+1=0x^2 + 2z^2 - y +1 = 0 x2+2z2−y+1=0
We get x2+2z2+1=yx^2 + 2z^2+1 = yx2+2z2+1=y.
For y=k, x2+2z2=k−1x^2 + 2z^2 = k-1x2+2z2=k−1.
This is a paraboloid.
Reference
- Multivariable_Calculus_8th_Edition (10.3, 12.5-12.6), James Stewart
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