Calculus:  The Language of Change

Exercise set 5.3.1-CD
Differential Equations and 
Functional Equations

by Brandon W. Buster

1. Computer Exercise

A view in the Microscope.  Sketch the view you would see in a powerful microscope focused on the graph:  y=Sin[θ] above the point where θ = (π/3).  Verify your prediction using the program Micro1D.

Clear[f, dθ, df, θ];
f[θ_] :=  Sin[θ];
df[dθ_]:=f'[θ] dθ;
Print["y = f[θ] = ", f[θ]];
θ = (π/3);
Print["VERY near θ = ", θ];
Print["Looks like dy = ", df[dθ]];
Plot[df[dθ], {dθ, -1, 1}, Ticks -> None, 
        AxesLabel -> {"dθ", "dy"}, 
        AspectRatio -> Automatic, PlotRange -> {-1, 1}, 
        PlotStyle -> {Red}];
scale  =  1;
Plot[{(f[θ + dθ] - f[θ]), df[dθ]}, {
          dθ, -scale, scale}, PlotRange -> {-scale, scale}, 
        PlotStyle -> {{Thickness[0.01], Blue}, {Red}}];

2. Derivative of Cosine.

Use a powerful microscope to prove that the differential of cosine is minus sine,
		y = Cos[θ]  =>  dy = -Sin[θ]dθ.

As we did in Figure CD-5.2, consider the apparent triangle similarity, comparing the long sides of the two triangles, 

		Sin[θ]/1 ≈ ((-δcos)/δ)θ = -\(TraditionalForm
      \`\(Cos[θ + δθ]\  - \ 
          Cos[θ]\)\/δθ\)

This gives us	ϵ = Sin[θ]/1 + \(TraditionalForm
      \`\(Cos[θ + δθ]\  - \ 
          Cos[θ]\)\/δθ\)

so		Cos[θ+δθ] - Cos[θ] = - Sin[θ] δθ + ϵ δθ

with ϵ ≈ 0 whenever δθ ≈ 0.  This has the form:

		f[θ + δθ] - f[θ] = f'[θ] + ϵ δθ

with f[θ] = Cos [θ] and f'[θ] = - Sin[θ].  Therefore, for θ in radians,

		y = Cos[θ]  =>  dy = -Sin[θ]dθ.
	
with ϵ ≈ 0 if δθ ≈ 0 (θ can take any value.)

3.

Part 1

Write an exact formula for Cos[x + δx] = Cos[x]Cos[δx]... using the addition formula for Cosine.  Compare the exact formula with the increment approximation obtained from the microscopic view of the circle.  What is the exact formula for ϵ?

Solution to Part 1.

Recall that the addition formula for cosine is:
	cos[x ± y] = cos[x] cos[y] ∓ sin[x] sin[y]

Thus, cos[x + δx] = cos[x] cos[δx] - sin[x] sin[δx]

The increment equation is:
	f[x + Δx] - f[x] = f'[x] Δx + ϵ Δx

let f[x] = Cos[x] and we get:
	cos[x + δx] - cos[x] = -sin[x]δx + ϵ δx

Solving this for cos[x + δx] we obtain:
	cos[x + δx] = -sin[x]δx + ϵ δx + cos[x]

Using what we have above we see:
	-sin[x]δx + ϵ δx + cos[x] = cos[x] cos[δx] - sin[x] sin[δx]

Solve this for ϵ we have:
	ϵ δx = cos[x] cos[δx] - sin[x] sin[δx] - cos[x] + sin[x]δx

ϵ =  \(TraditionalForm
      \`\(cos[x]\ cos[δx]\  - \ sin[x]\ sin[δx]\  - \ 
          cos[x]\)\/δx\) + sin[x]

Solving the increment equation for ϵ we get:
	ϵ = \(TraditionalForm
      \`\(f[x\  + \ Δx]\  - \ f[x]\)\/Δx\)- f'[x]

= \(TraditionalForm
      \`\(cos[x\  + \ δx]\  - \ cos[x]\)\/δx\) + sin[x]

We can use what we found above to get
	ϵ = \(TraditionalForm
      \`\(cos[x]\ cos[δx]\  - \ sin[x]\ sin[δx]\  - \ 
          cos[x]\)\/δx\) + sin[x]

Thus ϵ is the same for each equation!

Part 2.

Approximate the value of Cosine of 46 degrees using the linear increment approximation, discarding the ϵ δx term.  Give your answer in terms of exact constants such as π and (Sqrt[2]) as well as numerically.  Compare your approximation with the computer or your scientific calculator's approximation.

Solution to Part 2.

\(TextForm
    \`θ\  = \ 
      \(π\/4\ and\ \ \ δθ = \ π\/180 => 0\), 
    \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f[θ] = \ Cos[θ]\)\(TextForm
    \`f[θ\  + δθ]\  - \ f[θ]\  = \ 
      \(f'\)[θ]\  + \ ϵδθ\)\(TextForm
    \`Cos[θ\  + δθ] - \ Cos[θ]\  = \ 
      \(-Sin[θ]\) δθ\)\(TextForm
    \`Cos[π\/4 + \ π\/180] - Cos[π\/4] = 
      \(\(-Sin[π\/4]\)\ π\/180\  = \ 
        \(Cos[π\/4 + \ π\/180] - \ \@2\/2 = \ 
          \(-\((\@2\/2)\)\) π\/180\)\)\)\(TextForm\`\ 
    Cos[π\/4 + \ π\/180] = \ 
      \(\(-\((\@2\/2)\)\) π\/180\  + \ \@2\/2\  ≈ \ 
          \(-0.01234134\)\  + \  .7071067\  = \ O  .694765439\)
\)Calculator value for cos (46^o) is 0.6946583705

Part 3.

Use the increment approximation for Cosine to show that:

(lim_(Δθ  -> 0)) \(TraditionalForm
      \`\(Cos[π\/4\  + \ Δθ]\  - \ 
          Cos[π\/4]\)\/Δθ\) = -(1/(Sqrt[2]))

Solution to Part 3.

From Exercise Set 5.1 we know that f'[θ] = (lim_(Δθ  -> 0))\(TraditionalForm
      \`\(f[θ\  + \ Δθ]\  - \ 
          f[θ]\)\/Δθ\).

If f[θ] = Cos[θ] 	then f'[θ] = (lim_(Δθ  -> 0))\(TraditionalForm
      \`\(Cos[π\/4\  + \ Δθ]\  - \ 
          Cos[π\/4]\)\/Δθ\).

From part 1 above, we know that the increment approximation for Cosine is:

cos[x + δx] - cos[x] = -sin[x]δx + ϵ δx

Thus - Sin[θ] = f'[θ]

Moreover, -Sin[θ] = -(1/(Sqrt[2])).  Thus (lim_(Δθ  -> 0)) \(TraditionalForm
      \`\(Cos[π\/4\  + \ Δθ]\  - \ 
          Cos[π\/4]\)\/Δθ\) = -(1/(Sqrt[2])).