Practice Final Exam

MAT 229, Spring 2021

Consider the function y(x) defined implicitly by x=(Subscript[log, 2])(y).

Compute the derivative d y/d x of the function y(x).

Find the tangent line to the graph of y(x) at the point (1,2).

f[x_]:=2^x
f'[x]
Plot[{f[x],f'[1](x-1)+2},{x,0.5,1.5}]

Compute the integral using integration by parts, showing all the steps in your work.
	∫(tan^(-1))(2x)d x

Let u=(tan^(-1))(2x) so that d u=(1/(1+(((2x))^2)))2d x, and let d v=d x with v=x, then using integration by parts
	∫(tan^(-1))(2x)d x=x (tan^(-1))(2x)-∫((2x)/(1+4(x^2)))d x
In the new integral use the substitution u=1+4(x^2), d u=8x d x to get
	∫(tan^(-1))(2x)d x=x (tan^(-1))(2x)-∫((2x)/(1+4(x^2)))d x=x (tan^(-1))(2x)-∫(1/(4u))d u
	=x (tan^(-1))(2x)-(1/4)ln(1+4(x^2))+C

Evaluate ∫ (cos^3)(2x)(sin^3)(2x)d x  using an appropriate trig substitution.

f[x_]=Integrate[Cos[2x]^3Sin[2x]^3,x]
g[u_]=Integrate[(u^3-u^5)(-1/2),u]
Plot[{f[x],g[Cos[2x]]},{x,0,Pi}]

(Power series) Consider the function given as the power series centered at 0.
	f(x)=(((Underoverscript[Σ, (n=0), ∞])(-1))^n)((x^(n+2))/((n+1)(3^n)))

What is the domain for f(x)? In other words, what is the interval of convergence?

Using the ratio test,
	(lim_(n->∞))|((((-1))^(n+1))((x^((n+1)+2))/((n+1+1)(3^(n+1)))))/((((-1))^n)((x^(n+2))/((n+1)(3^n))))|=(lim_(n->∞))((n+1)/(n+2))((|x|)/3)=((|x|)/3).
By the ratio test, the series converges if ((|x|)/3)<1 or -3<x<3. Check the endpoints using another test.
x=-3: (((Underoverscript[Σ, (n=0), ∞])(-1))^n)((((-3))^(n+2))/((n+1)(3^n)))=(((Underoverscript[Σ, (n=0), ∞])(-1))^(2n+2))(9/(n+1))=9(Underoverscript[Σ, (n=0), ∞])(1/(n+1))=9(Underoverscript[Σ, (n=1), ∞])(1/n)
	This diverges since it is a p-series with p=1, the harmonic series.
x=3: (((Underoverscript[Σ, (n=0), ∞])(-1))^n)((((3))^(n+2))/((n+1)(3^n)))=(((Underoverscript[Σ, (n=0), ∞])(-1))^n)(9/(n+1))
	This is an alternating series that satisfies the alternating series test hypothesis. It converges.

The domain is (-3,3].

Find a power series centered at 0 for f'(x) and give its radius of convergence.

f'(x)=(((((Underoverscript[Σ, (n=0), ∞])(-1))^n)((x^(n+2))/((n+1)(3^n)))))'=(((Underoverscript[Σ, (n=0), ∞])(-1))^n)(((n+2)(x^(n+1)))/((n+1)(3^n)))
This will have the same radius of convergence as the series for f(x), namely 3.

We wish to approximate the function f(x)=ln(x) as a Taylor polynomial of degree 3, (Subscript[T, 3])(x), in the vicinity of x=1.

Write the polynomial.

Use the Taylor remainder theorem to give a good bound on the error of our approximation on the interval [0.5, 1.5].  A graph of the fourth derivative of f(x) on this interval is provided here:

t3[x_]=Normal[Series[Log[x],{x,1,3}]]
Plot[Log''''[x],{x,0.5,1.5},PlotRange->All]
errorbound=100/4!(0.5)^4
Plot[ {t3[x]-Log[x]},{x,0.5,1.5},PlotRange->All]

Consider the integral (∫_0^2)(1/(1+x))d x.

Approximate it using the midpoint rule with 2 rectangles.

(∫_0^2)(1/(1+x))d x≈(1/(1+1/2))+(1/(1+3/2))=(2/3)+(2/5)=(16/15)≈1.067

What does the midpoint rule error estimate give as the maximum error in this approximation?

The second derivative of the integrand is
	(((1/(1+x))))''=(((((1+x))^(-1))))'=((-1(((1+x))^(-2))))'=2(((1+x))^(-3))=(2/(((1+x))^3))
The maximum value of this second derivative on [0,2] is achieved at the left endpoint x=0. Use M=2.
	error≤(2/(24(((2))^2)))(((2-0))^3)=(4/24)=(1/6)≈0.1667

Two of the following three series can be evaluated almost instantaneously; the third is divergent. Give their values, or prove divergence.

(Underoverscript[Σ, (n=0), ∞])((2^(n+1))/(((-3))^n))

(Underoverscript[Σ, (n=0), ∞])((2^n)/(n!))

(Underoverscript[Σ, (n=0), ∞])((1+(1/n))/((n+1)))

Sum[2^(n+1)/(-3)^n,{n,0,Infinity}]
2(1/(1+2/3))
Sum[2^n/n!,{n,0,Infinity}]
E^2
Sum[(1+1/n)/(n+1),{n,0,Infinity}]
(* Hopefully a limit comparison with 1/n: *)
Limit[((1+1/n)/(n+1))/(1/n),n->Infinity]

Determine whether the improper integral (∫_e^∞)(1/(x ln(x)))d x converges or not. Give reasons for your answer.

Using substitution with u=ln(x), d u=(1/x)d x:
	(∫_e^∞)(1/(x ln(x)))d x=(lim_(R->∞))(∫_e^R)(1/(x ln(x)))d x=(lim_(R->∞))(∫_(x=e)^(x=R))(1/u)d u=(lim_(R->∞))ln(ln(x))(Subscript[|, x=e]^(x=R))=∞
This improper integral diverges.

This is the graph  of the beautiful “rose function”, r=sin(2 θ).

What is the period of this function?

Give a choice of angle values θ that trace out just the dark petal.

Write an integral representing the area of one petal, and compute that area (including a decimal value).

r[n_,θ_]:=Sin[n θ]
Show[
PolarPlot[{r[2,θ]},{θ,0,2Pi},PlotStyle->Orange],
PolarPlot[{r[2,θ]},{θ,0,Pi/2},PlotStyle->Black]
]
Integrate[1/2 r[2,θ]^2,{θ,0,Pi/2}]
Integrate[1/4(1-Cos[4θ]),{θ,0,Pi/2}]
N[%]

The curve given parametrically by x=cos(2t),y=sin(3t) is shown below.  It has two points with horizontal tangents. Determine the coordinates (x,y) for those two points.

ParametricPlot[{Cos[2t],Sin[3t]},{t,0,4π},Ticks->None,AxesLabel->{x,y},BaseStyle->FontSize->14]

Solve ((d y)/(d x))=0. Since ((d y)/(d x))=((y'(t))/(x'(t))), the slope is 0 when y'(t)=0. Solve
	((sin(3t)))'=0
⟶3cos(3t)=0
⟶cos(3t)=0
⟶3t=(π/2)+k π
⟶t=(π/6)+k (π/3)
The point in the first quadrant is (x,y)=(cos(2(π/6)),sin(3(π/6)))=((1/2),1). By symmetry, the one in the fourth quadrant is ((1/2),-1).

Consider the vector  (Overscript[v, ->])=⟨-1,3,2⟩

Compute the exact length of  (Overscript[v, ->]) (that is, the norm of  (Overscript[v, ->]), or | (Overscript[v, ->])|).

Find a vector (Overscript[u, ->]) perpendicular to  (Overscript[v, ->]) (and show that it is perpendicular).

Use the cross product to find a vector perpendicular to both  (Overscript[u, ->]) and  (Overscript[v, ->]).

v={-1,3,2}
Norm[v]
u={3,1,0}
u.v
Cross[u,v]

Consider the plane through the three points (Subscript[P, 0])(1,0,0), (Subscript[P, 1])(1,1,0), (Subscript[P, 2])(0,1,1). Find an equation for this plane.

Two vectors parallel to this plane are (Overscript[(Subscript[P, 1])(Subscript[P, 0]), ->])=⟨0,1,0⟩ and (Overscript[(Subscript[P, 2])(Subscript[P, 0]), ->])=⟨-1,1,1⟩. A normal vector is
	(Overscript[(Subscript[P, 1])(Subscript[P, 0]), ->])×(Overscript[(Subscript[P, 2])(Subscript[P, 0]), ->])=|({{i, j, k}, {0, 1, 0}, {-1, 1, 1}})|=i-0j+k=⟨1,0,1⟩
Using the first point, an equation for the plane is
	(x-1)+z=0
or
	x+z=1