{NotebookFileName[],DateString[]}

Screwing around

style[
t_
]:=
Style[t,Bold,FontFamily->"Arial",FontSize->14]

D[x Log[x]-x,x]

Series[x!,{x,∞,5}]

D[(Sqrt[3x+(x^2)]),x]

Factor[(x^5)-2x+2]

D[(x^4)-((y[x])^3)-y[x],x]

Solve[0==D[(x^4)-((y[x])^3)-y[x],x],y'[x]]

y'[x]/.Solve[0==D[(x^4)-((y[x])^3)-y[x],x],y'[x]][[1]]

y'[x]/.Solve[0==D[(x^4)-((y[x])^3)-y[x],x],y'[x]][[1]]/.{x->1,y[x]->0}

Dt[(x^4)-(y^3)-y,x]

D[ArcCos[x],x]

D[Cos[x]+x Sin[y[x]]-y[x],x]

Solve[0==D[Cos[x]+x Sin[y[x]]-y[x],x],y'[x]]

Slow-pitch softball

According to this site: https://www.sportsknowhow.com/softball/dimensions/softball-slow-pitch-field-dimensions.html, the typical distance between bases is 60 ft. Assuming the pitcher’s mound is in the center, it’s distance from the home plate would be:

(60/(Sqrt[2]))//N

The same site also gives the typical distance from the pitcher’s mound to Home Plate as 35-53 ft. Taking the average:

((35+53)/2)

So, they’re pretty consistent. Converting to meters:

UnitConvert[Quantity[42.0,"Feet"]]

So, call it 13 m.

Assume the ball starts and ends at the same height, and that at the midpoint of the arc its height is a greater than that (a for arc height). Then, if the pitch takes time t from pitcher to home, the vertical fall of the ball is a=(1/2)((g((t/2)))^2), since it falls a in the second half of the pitch. Then that gives me t as a function of a and thence v, the velocity of the pitch, as a function of a.

Solve[a==(1/2)g((((t/2)))^2),t]

QuantityMagnitude[UnitConvert[Quantity[1,"StandardAccelerationOfGravity"],(("Meters")/(("Seconds")^2))]]//N

v1[
a_,
pd_:13.0,
g_:QuantityMagnitude[UnitConvert[Quantity[1,"StandardAccelerationOfGravity"],(("Meters")/(("Seconds")^2))]]
]:=Module[{t,v},
t=2(Sqrt[((2a)/g)]);
v=pd/t;
v
]

v1[1]

UnitConvert[Quantity[v1[1],(("Meters")/("Seconds"))],(("Miles")/("Hours"))]

v1[0.5]

v1[0.4]

v1[2]

v1[3]

v1[1.9]

v1[0.1]

v1[a]

The rules about the height of the arc seem to be very vague. One document (https://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwjnp6rg5ebWAhXm7IMKHXNkAh4QFggoMAA&url=http%3A%2F%2Fpsblions.com%2Finformation%2Fintramurals-facilities%2FIntramural_Website%2FRules%2Frulessoftball.doc&usg=AOvVaw1OxeryaNTxMOW7Q1ujUR1W) says “The ball must be delivered underhand at a moderate speed from below the hip, and with a perceptible arc that reaches 6 feet from the ground but not exceeding 12 feet.  The speed of the pitch is left entirely to the judgement of the umpire.” Assuming an underhand pitch starts at about 3 ft, this would make the minimum arc 3 ft, or about 1 m, and the maximum 9 ft, or about 3 m.

D[v1[a],a]

D[v1[a],a]/.{{a->1},{a->2},{a->3}}

Block[{is=400,ar=1,s=1,m=2,a,v,vy0,g,pd=13,tmax},
g=QuantityMagnitude[UnitConvert[Quantity[1,"StandardAccelerationOfGravity"],(("Meters")/(("Seconds")^2))]];
a=m-s;
v=v1[a];
tmax=(pd/v);
h1=(t|->{v t,a+(1/2)(g tmax t-g (t^2))});
{
ParametricPlot[h1[t],{t,0,tmax}
,PlotRange->{{0,1.2pd},{0,All}}
,Axes->None
(*,PlotLegends->{"ball trajectory"}*)
,Frame->{{True,False},{True,False}}
,FrameStyle->Directive[Bold,FontFamily->"Arial",FontSize->14]
,FrameLabel->{"distance from pitcher's mound (m)","height (m)"}
,ImageSize->is
,AspectRatio->ar
,Epilog->{
Thick,PointSize[Large],Arrowheads[{-Automatic,Automatic}],
Arrow[{{0,s},{pd,s}}],Text[style["t"],{(pd/2),s},{0,1}],
Arrow[{{0,s+a},{(pd/2),s+a}}],Text[style["t/2"],{(pd/4),s+a},{0,-1}],
Arrow[{{(pd/2),s+a},{pd,s+a}}],Text[style["t/2"],{((3pd)/4),s+a},{0,-1}],
Arrow[{{(pd/2),s},{(pd/2),s+a}}],Text[style["x"],{(pd/2),s+(a/2)},{-2,0}],
Arrow[{{pd,s},{pd,s+a}}],Text[style["(1/2)g((t/2))^2"],{pd,s+(a/2)},{-1,0}]
}
],
ParametricPlot[h1[t],{t,0,tmax}
,PlotRange->{{0,1.2pd},{0,1.4(s+a)}}
,Axes->None
(*,PlotLegends->{"ball trajectory"}*)
,Frame->{{True,False},{True,False}}
,FrameStyle->Directive[Bold,FontFamily->"Arial",FontSize->14]
,FrameLabel->{"distance from pitcher's mound (m)","height (m)"}
,ImageSize->is
,AspectRatio->Automatic
,Epilog->{
Thick,PointSize[Large],Arrowheads[{-Automatic,Automatic}],
Arrow[{{0,s},{pd,s}}],Text[style["t"],{(pd/2),s},{0,1}],
Arrow[{{0,s+a},{(pd/2),s+a}}],Text[style["t/2"],{(pd/4),s+a},{0,-1}],
Arrow[{{(pd/2),s+a},{pd,s+a}}],Text[style["t/2"],{((3pd)/4),s+a},{0,-1}],
Arrow[{{(pd/2),s},{(pd/2),s+a}}],Text[style["x"],{(pd/2),s+(a/2)},{-2,0}],
Arrow[{{pd,s},{pd,s+a}}],Text[style["(1/2)g((t/2))^2"],{pd,s+(a/2)},{-1,0}]
}
]
}
]//Column

D[(S/(Sqrt[x])),x]

D[v1[x],x]

Plot[Evaluate[D[v1[x],x]],{x,1,3}
,PlotRange->{{1,3},{All,0}}
,Axes->None
,Frame->{{True,False},{False,True}}
,FrameTicks->All
,LabelStyle->Directive[Bold,FontFamily->"Arial",FontSize->14]
,FrameLabel->{{ "sensitivity ((m/s)/m)",None},{None,"arc height (m)"}}
,ImageSize->400
]

Solve[D[v1[x],x]==-5,x]

Examples

Factor[(x^2)+3x-40]