Review for “New Stuff” final

Exam rules

You can have one page of notes (front and back) and you are allowed to have a calculator.

You must show full work on all integrals.

Topics

Parametric equations

Be familiar with techniques for changing parametric equations x=x(t), y=(t) into a single Cartesian equation.

Find parametric equations for circles.

Calculus with parametric equations

Slopes and tangent lines.

Horizontal and vertical tangents.

Length of curves parametrically defined.

Polar coordinates

Convert between Cartesian coordinates and polar coordinates.

Knowing the simpler equations in polar coordinates.

Polar curves r=f(θ).

Writing polar curves as parametric equations to find slopes and lengths.

Finding periods of polar curves.

Areas of polar curves

Differences of areas to find area between curves.

Finding intersections of curves.

3D coordinates

Distance formula.

Equations for simple objects (spheres, planes parallel to coordinates axes, etc.).

Vectors

Significance of vector components.

Magnitude.

Direction of 2D vectors as an angle with the positive x-axis.

Scalar multiplication and its geometric significance.

Vector addition and its geometric significance.

Unit vectors i, j, k. Orthogonal coordinate systems. Frenet Frame.

Dot product

Find angle between two vectors.

Test if two vectors are perpendicular to each other.

Dot products relationship to magnitude.

Vector projections.

Cross product

Only defined for 3d space vectors.

(Overscript[u, ->])×(Overscript[v, ->]) is perpendicular to both (Overscript[u, ->]) and (Overscript[v, ->]).

(Overscript[u, ->])×(Overscript[v, ->]) is a good test of when vectors (Overscript[u, ->]) and (Overscript[v, ->]) are parallel (the cross product is the 0 vector).

|(Overscript[u, ->])×(Overscript[v, ->])| is the area of the parallelogram formed from  (Overscript[u, ->]) and (Overscript[v, ->]).

|(Overscript[u, ->])×(Overscript[v, ->])| is twice the area of the triangle formed from  (Overscript[u, ->]) and (Overscript[v, ->]).

Derivation of Kepler’s First Law.

Studying

Review your handouts and the problems we did together found therein.

Try problems you haven’t worked before from the exercises in your reference textbook.

Sample questions

Consider the parametric equations x=4sin(π t),y=4cos(π t)+1.

Find an equation for its tangent line at t=1/4.

A point on this line is (x,y)=(4sin(π/4),4cos(π/4)+1)=(2(Sqrt[2]),1+2(Sqrt[2]))≈(2.83,3.83).
The slope for the tangent line is ((d y)/(d x))(Subscript[|, t=1/4])=((d y/d t)/(d x/d t))(Subscript[|, t=1/4])=((-4π sin(π t))/(4π cos(π t)))(Subscript[|, t=1/4])=((- sin(π/4) )/(cos(π/4)))=-1.
An equation for the tangent line is
	y-(1+2(Sqrt[2]))=-(x-2(Sqrt[2])) 	or	y=-x+4(Sqrt[2])+1	or	y=-x+6.66

Find all points on the curve that have horizontal tangents.

A point on this line with a horizontal tangent satisfies ((d y)/(d x))=0 or
	-4π sin(π t)=0
		⟶	sin(π t)=0
		⟶	t any integer
The corresponding points on the curve are
	(x,y)=(4 0,4 1+1)=(0,5) for t even
	(x,y)=(4 0,4×(-1)+1)=(0,-3) for t odd

Eliminate the parameter to find a Cartesian equation of the curve.

First solve the x equation for sin(π t) and the y equation for cos(π t). Then use the Pythagorean identity (cos^2)(θ)+(sin^2)(θ)=1 to get
	(((((y-1)/4)))^2)+((((x/4)))^2)=1	or	(x^2)+(((y-1))^2)=16

Polar coordinates for a point are (r,θ)=(4,π/3).

Plot the point in the plane.

Graphics[{Blue, PointSize[0.03], Point[4 {Cos[Pi/3], Sin[Pi/3]}], Dashing[{0.02}], Line[{{0, 0}, 4 {Cos[Pi/3], Sin[Pi/3]}}], Circle[{0, 0}, .3, {0, Pi/3}], Text[Style[4, FontSize -> 14], 2{Cos[Pi/3], Sin[Pi/3]}, {1, -1}], Text[Style[π/3, FontSize -> 14], .4 {Cos[Pi/6], Sin[Pi/6]}]}, Axes -> True, AxesStyle -> Blue, AspectRatio -> Automatic]

Find two other different polar coordinates for this same point, one with r<0 and one with r>0 but a different value for θ.

To get polar coordinates for this point with r<0 use r=-4 and θ=π/3+π=4π/3. (Or any 2π multiple added to this.)
	(r,θ)=(-4,4π/3)

To get polar coordinates for this point with r>0 and a different angle use  any 2π multiple to θ.
	(r,θ)=(4,π/3+2π)=(4,7π/3)

Give Cartesian coordinates (x,y) for this point.

Use the equations relating Cartesian coordinates to their polar coordinate equivalents.
	x=r cos(θ)=4cos(π/3)=4×(1/2)=2.
	y=r sin(θ)=4sin(π/3)=4 ((Sqrt[3])/2)=2(Sqrt[3]).
Cartesian coordinates for this point are (x,y)=(2,2(Sqrt[3]))≈(2,3.5).

What is the area enclosed by the polar curve r=(Sqrt[cos(θ)+1]), 0≤θ≤2π, shown below.

PolarPlot[(Sqrt[Cos[θ]+1]),{θ,0,2π}]

Use the polar area formula
	area=(∫_0^(2π))(1/2)(r^2)d θ=(∫_0^(2π))(1/2)((((Sqrt[cos(θ)+1])))^2)d θ=(1/2)(∫_0^(2π))cos(θ)+1d θ
	=(1/2)(sin(θ)+θ)(Subscript[|, 0]^(2π))=(1/2)(sin(2π)+2π)-(1/2)(sin(0)+0)=π

Find the slope of the tangent line to the polar curve r=θ when θ=π/4.

Converting the polar curve to parametric equations using x=r cos(θ),y=r sin(θ) produces
	x=θ cos(θ)
y=θ sin(θ)
The slope of the tangent line is
	(Subscript[(Subscript[(Subscript[((d y)/(d x))|, θ=π/4])=((d y/d θ)/(d x/d θ))|, θ=π/4])=((sin(θ)+θ cos(θ))/(cos(θ)-θ sin(θ)))|, θ=π/4])
	=((sin(π/4)+π/4 cos(π/4))/(cos(π/4)-π/4 sin(π/4)))=(((Sqrt[2])/2+π/4((Sqrt[2])/2))/((Sqrt[2])/2-π/4((Sqrt[2])/2)))=((4+π)/(4-π))≈8.31958

Find a Cartesian equation for the curve given by r=2cos(θ).

Multiply the polar equation by r to get
	(r^2)=2r cos(θ)
Use the equations relating Cartesian coordinates and polar coordinates to rewrite this in Cartesian coordinates as
	(x^2)+(y^2)=2x

Find an equation for the sphere with center (2,-3,6) that just touches the x-y plane.

If it just touches the x-y plane, its radius is the distance from the center to that plane which is 6. The equation is 
	(((x-2))^2)+(((y+3))^2)+(((z-6))^2)=36.

Find a unit vector that points in the same direction as the vector that points from point (1,0,-2) to point (3,1,1).

The vector that points from point (1,0,-2) to point (3,1,1) has components ⟨3-1,1-0,1-(-2)⟩=⟨2,1,3⟩. A unit vector that points in the same direction is (1/(Sqrt[(2^2)+(1^2)+(3^2)]))⟨2,1,3⟩=(1/(Sqrt[14]))⟨2,1,3⟩=⟨(2/(Sqrt[14])),(1/(Sqrt[14])),(3/(Sqrt[14]))⟩.

Find the values of x such that the angle between the vectors ⟨2,1,-1⟩ and ⟨1,x,0⟩ is 45°.

The angle between these two vectors is given by
	cos(θ)=((⟨2,1,-1⟩·⟨1,x,0⟩)/(|⟨2,1,-1⟩||⟨1,x,0⟩|))=((2+x)/((Sqrt[6])(Sqrt[1+(x^2)])))
If θ=45°, then
	((Sqrt[2])/2)=((2+x)/((Sqrt[6])(Sqrt[1+(x^2)])))
⟶(Sqrt[12])(Sqrt[1+(x^2)])=4+2x
⟶((((Sqrt[12])(Sqrt[1+(x^2)])))^2)=(((4+2x))^2)
⟶12(1+(x^2))=16+16x+4(x^2)
⟶8(x^2)-16x-4=0
⟶x=((16±(Sqrt[256+128]))/16)=((16±(Sqrt[384]))/16)=((2±(Sqrt[6]))/2)
There are two such values of x.

Consider the two vectors (Overscript[u, ->])=(Overscript[i, ->])-(Overscript[j, ->])+(Overscript[k, ->])  and (Overscript[v, ->])=(Overscript[-2i, ->])+3(Overscript[k, ->]).

What is the area of the parallelogram whose sides are formed from  (Overscript[u, ->]) and (Overscript[v, ->])?

The area of the parallelogram is |(Overscript[u, ->])×(Overscript[v, ->])|. Compute the cross product.
	(Overscript[u, ->])×(Overscript[v, ->])=det(({{(Overscript[i, ->]), (Overscript[j, ->]), (Overscript[k, ->])}, {1, -1, 1}, {-2, 0, 3}}))=(Overscript[i, ->])(-3-0)-(Overscript[j, ->])(3-(-2))+(Overscript[k, ->])(0-2)=-3(Overscript[i, ->])-5(Overscript[j, ->])-2(Overscript[k, ->])
The area of the parallelogram is |-3(Overscript[i, ->])-5(Overscript[j, ->])-2(Overscript[k, ->])|=(Sqrt[38]).

Find two different unit vectors that are orthogonal to (Overscript[u, ->]) and (Overscript[v, ->]).

An orthogonal vector is the cross product
	(Overscript[u, ->])×(Overscript[v, ->])=-3(Overscript[i, ->])-5(Overscript[j, ->])-2(Overscript[k, ->])
Two unit vectors parallel to this vector are
	±(1/(Sqrt[9+25+4]))(-3(Overscript[i, ->])-5(Overscript[j, ->])-2(Overscript[k, ->]))=∓(3/(Sqrt[38]))(Overscript[i, ->])∓(5/(Sqrt[38]))(Overscript[j, ->])∓(2/(Sqrt[38]))(Overscript[k, ->])