Two Words of Caution
3.2 Differential Calculus
3.2.4 Tangent Lines
If f%aexists, we interpret f%ato be the slope of the line tangent to the graph of y fxat the pointa, fa. An equation of the tangent is given by
yfa f%axa or y f%axa fa.
EXAMPLE 3.2.6: Find an equation of the line tangent to the graph of fx sinx1/3cos1/3x
at the point withx-coordinatex5Π/3.
SOLUTION: Because we will be graphing a function involving odd roots of negative numbers, we begin by loading theRealOnlypackage contained in the Miscellaneousfolder (or directory). We then define
fxand compute f%x.
In[289]:= << Miscellaneous‘RealOnly‘
In[290]:= fx Sinxˆ1/3 Cosxˆ1/3 In[291]:= fx
Out[291]= Cosx1/3
3 x2/3 Sinx 3 Cosx2/3
Then, the slope of the line tangent to the graph offxat the point with
while they-coordinate of the point is In[293]:= f5Π/3
Thus, an equation of the line tangent to the graph of fxat the point withx-coordinatex5Π/3is as shown in Figure 3-5.
In[295]:= p1 Plotfx,x, 0, 4Π, DisplayFunctionIdentity p2 ListPlot5Π/3, f5Π/3//N,
PlotStylePointSize0.03, DisplayFunctionIdentity
p3 Plotf5Π/3x5Π/3 f5Π/3,x, 0, 4Π, PlotStyle> GrayLevel0.6,
DisplayFunctionIdentity
In[296]:= Showp1, p2, p3, AspectRatio> Automatic, DisplayFunction$DisplayFunction
EXAMPLE 3.2.7: Find an equation of the line tangent to the graph of fx 94x2at the point1, f1.
2 4 6 8 10 12 1
2 3 4 5
Figure 3-5 fx sinx1/3cos1/3xtogether with its tangent at the point5Π/3, f5Π/3
-3 -2 -1 1 2 3
-10 -7.5 -5 -2.5 2.5 5 7.5 10
Figure 3-6 fxtogether with its tangent at1, f1
SOLUTION: After definingf, we see that f1 5and f%1 8 In[297]:= fx 94xˆ2
f1 f1 Out[297]= 5 Out[297]= 8
so an equation of the line tangent toy fxat the point1,5isy5 8x1ory 8x13. We can visualize the tangent at1, f1with Plot. (See Figure 3-6.)
In[298]:= Plotfx, f1x1 f1,x,3, 3, PlotStyle>GrayLevel0, GrayLevel0.3, PlotRange>10, 10
Figure 3-7 An animation
In addition, we can view a sequence of lines tangent to the graph of a function for a sequence ofxvalues usingDo. In the following, we use Doto generate graphs ofy fxandy f%axafafor 50 equally spaced values ofabetween3and3. (See Figure 3-7.)
In[299]:= DoPlotfx, faxa fa,x,3, 3, PlotStyle>GrayLevel0, GrayLevel0.3, PlotRange>10, 10,a,2, 2, 4/49
On the other hand,
In[300]:= listofgraphics Table
Plotfx, faxa fa,x,3, 3, PlotStyle>GrayLevel0, GrayLevel0.3, PlotRange>10, 10,
DisplayFunction> Identity,a,2, 2, 4/8 toshow Partitionlistofgraphics, 3
ShowGraphicsArraytoshow
graphsy fxandy f%axa fafor nine equally spaced values ofabetween3and3and displays the result as a graphics array. (See Figure 3-8.)
In the graphs, notice that where the tangent lines have positive slope (f%x> 0), fxis increasing while where the tangent lines have nega-tive slope (f%x<0), fxis decreasing.
-3 -2 -1 1 2 3
Figure 3-8 fxtogether with various tangents
Tangent Lines of Implicit Functions
EXAMPLE 3.2.8: Find equations of the tangent line and normal line to the graph ofx2yy3 8 at the point3,1. Find and simplifyy%%
d2y/ dx2.
SOLUTION: We will evaluate y% dy/ dx if x 3 and y 1 to determine the slope of the tangent line at the point3,1. Note that we cannot (easily) solvex2yy38foryso we use implicit
In[302]:= s1 Dteq
Out[302]= 2 x y Dtx x2Dty 3 y2Dty 0 In[303]:= s2 s1/.Dtx 1
Out[303]= 2 x yx2Dty 3 y2Dty 0 In[304]:= s3 Solves2, Dty
Out[304]= Dty 2 x y x23 y2
Notice thats3is alist.The formula fory%dy/ dxis the second part of
Listsare discussed in more
detail in Chapter 4. the first part of the first part ofs3and extracted froms3with
In[305]:= s31, 1, 2 Out[305]= 2 x y
x23 y2
We then useReplaceAll(/.) to find that the slope of the tangent at 3,1is
In[306]:= s31, 1, 2/.x 3, y1 Out[306]= 1
The slope of the normal is1/1 1. Equations of the tangent and normal are given by
y11x3 and y1 1x3,
respectively. See Figure 3-9.
In[307]:= cp1 ContourPlotxˆ2yyˆ38,x,5, 5,y,5, 5, Contours 0, ContourShadingFalse,
PlotPoints200, DisplayFunctionIdentity p1 ListPlot3, 1,
PlotStylePointSize0.03, DisplayFunctionIdentity
p2 Plotx3 1,x3 1,x,5, 5, PlotStyleGrayLevel0.3,
DisplayFunctionIdentity
Showcp1, p1, p2, FrameFalse, AxesAutomatic, AxesOrigin 0, 0, AspectRatioAutomatic, DisplayFunction$DisplayFunction
To findy%%d2y/ dx2, we proceed as follows.
In[308]:= s4 Dts31, 1, 2//Simplify
-4 -2 2 4
-5 -2.5 2.5 5 7.5
Figure 3-9 Graphs ofx2yy38(in black) and the tangent and normal at3,1(in gray)
Out[308]= 2x23 y2 y Dtx x Dty x23 y22
In[309]:= s5 s4/.Dtx 1/.s31//Simplify Out[309]= 6 yx2y2 x23 y2
x23 y23 The result means that
y%%d2y
dx2 6x2yy3 x23y2 x23y23 .
Becausex2yy38, the second derivative is further simplified to y%% d2y
dx2 48x23y2 x23y23 .
Parametric Equations and Polar Coordinates For the parametric equationsx ft, ygt,t'I,
y%dy
dx dy/ dt dx/ dt g%t
f%t and
y%% d2y dx2 d
dx dy
dx d/ dtdy/ dx dx/ dt .
Ifx ft, ygthas a tangent line at the pointfa, ga, parametric equations of the tangent are given by
x fa t f%a and yga tg%a. (3.2)
If f%a,g%a 0, we can eliminate the parameter from (3.2) xfa
f%a yga g%a yga g%a
f%axfa
and obtain an equation of the tangent line in point-slope form.
In[310]:= l Solvexa t xa cx, t r Solveya t ya cy, t
Out[310]= BoxDatat cxxa xa Out[310]= BoxDatat cyya
ya
EXAMPLE 3.2.9 (The Cycloid): Thecycloidhas parametric equations xtsint and y1cost.
Graph the cycloid together with the line tangent to the graph of the cycloid at the pointxa, yafor various values ofabetween2Πand 4Π.
SOLUTION: After definingx and ywe use’ to computedy/ dt and dx/ dt. We then computedy/ dx dy/ dt/dx/ dtandd2y/ dx2.
In[311]:= xt tSint yt 1Cost dx xt
dy yt dydx dy/dx Out[311]= 1Cost Out[311]= Sint Out[311]= Sint
1Cost
In[312]:= dypdt SimplifyDdydx, t Out[312]= 1
1Cost
In[313]:= secondderiv Simplifydypdt/dx Out[313]= 1
1Cost2
We then useParametricPlotto graph the cycloid for2Π t 4Π, naming the resulting graphp1.
In[314]:= p1 ParametricPlotxt, yt,t,2Π, 4Π,
PlotStyle>GrayLevel0, Thickness0.01, DisplayFunction> Identity
Next, we useTableto definetoplotto be 40 tangent lines (3.2) using equally spaced values ofa between2Πand 4Π. We then graph each line toplotand name the resulting graph p2. Finally, we show p1 andp2together with theShowfunction. The resulting plot is shown to scale because the lengths of thexandy-axes are equal and we include the option AspectRatio->1. In the graphs, notice that on intervals for whichdy/ dxis defined,dy/ dxis a decreasing function and, conse-quently,d2y/ dx2 <0. (See Figure 3-10.)
In[315]:= toplot Table
xa t xa, ya t ya,a,2Π, 4Π, 6Π/39 p2 ParametricPlotEvaluatetoplot,
t,2, 2, PlotStyle> GrayLevel0.5, DisplayFunction> Identity
Showp1, p2, AspectRatio> 1, PlotRange>3Π, 3Π, DisplayFunction> $DisplayFunction
EXAMPLE 3.2.10 (Orthogonal Curves): Two linesL1andL2with slopes m1 and m2, respectively, are orthogonal if their slopes are negative reciprocals:m1 1/ m2.
Extended to curves, we say that the curvesC1andC2areorthogonal at a point of intersection if their respective tangent lines to the curves at that point are orthogonal.
Show that the family of curves with equationx22xyy2 C is orthogonal to the family of curves with equationy22xyx2C.
SOLUTION: We begin by definingeq1and eq2to be the left-hand sides of the equationsx22xyy2Candy22xyx2 C, respectively.
In[316]:= eq1 xˆ22x yyˆ2 eq2 yˆ22x yxˆ2
-5 5 10
-7.5 -5 -2.5 2.5 5 7.5
Figure 3-10 The cycloid with various tangents
We then useDtto differentiate andSolveto findy% dy/ dx. Because the derivatives are negative reciprocals, we conclude that the curves are orthogonal. We confirm this graphically by graphing several members of each family withContourPlotand showing the results together.
(See Figure 3-11.)
In[317]:= BoxDatas1 Dteq1/.Dtx> 1, Dty> dydx, Solves1 0, dydx
Out[317]= 2 x2 dydx x2 y2 dydx y Out[317]= dydx xy
xy
In[318]:= BoxDatas2 Dteq2/.Dtx> 1, Dty> dydx, Solves2 0, dydx
Out[318]= 2 x2 dydx x2 y2 dydx y Out[318]= dydx xy
xy
-4 -2 2 4
-4 -2 2 4
Figure 3-11 x22xyy2Candy22xyx2Cfor various values ofC
In[319]:= cp1 ContourPloteq1,x,5, 5,y,5, 5, ContourShading> False,
ContourStyle> GrayLevel0, Frame> False, Axes> Automatic, AxesOrigin>0, 0,
DisplayFunction> Identity, PlotPoints> 60 cp2 ContourPloteq2,x,5, 5,y,5, 5,
ContourShading> False,
ContourStyle> GrayLevel0.4, Frame> False, Axes> Automatic, AxesOrigin>0, 0,
DisplayFunction> Identity, PlotPoints> 60 Showcp1, cp2, DisplayFunction> $DisplayFunction
EXAMPLE 3.2.11 (Theorem 1. The Mean-Value Theorem for Deriva-tives): Ify fxis continuous ona, band differentiable ona, bthen there is at least one value ofcbetweenaandbfor which
f%c fb fa
ba or, equivalently, fb fa f%cba. (3.3) Find all number(s)cthat satisfy the conclusion of the Mean-Value Theorem for fx x23xon the interval0,7/2.
SOLUTION: By the power rule, f%x 2x3. The slope of the secant containing0, f0and7/2, f7/2is
f7/2 f0 7/20 1
2. Solving2x31/2forxgives usx7/4.
In[320]:= fx xˆ23x Out[320]= 3 xx2
In[321]:= Solvefx 0, x Out[321]= x 3
2
In[322]:= Solvefx f7/2 f0/7/20 Out[322]= x 7
4
x 7/4satisfies the conclusion of the Mean-Value Theorem for fx x23xon the interval0,7/2, as shown in Figure 3-12.
In[323]:= p1 Plotfx,x,2, 4,
DisplayFunctionIdentity p2 Plotfx,x, 0, 7/2,
PlotStyleThickness0.02, DisplayFunctionIdentity p3 ListPlot0, f0,7/4, f7/4,
7/2, f7/2,
PlotStylePointSize0.05, DisplayFunctionIdentity p4 Plotf7/4x7/4 f7/4,
f7/2 f0/7/20x,
x,2, 4, PlotStyle Dashing0.01, Dashing0.02,
DisplayFunctionIdentity Showp1, p2, p3, p4, DisplayFunction
$DisplayFunction, AspectRatioAutomatic
-2 -1 1 2 3 4
-4 -2 2 4
Figure 3-12 Graphs of fx x23x, the secant containing0, f0and7/2, f7/2, and the tangent at7/4, f7/4