#2, 8, 43, pp. 675--

#2

x[t_]:=1/t
y[t_]:=Sqrt[t]*Exp[-t]

Here’s the derivative dy/dx:

dydx[t_]=Simplify[y'[t]/x'[t]]

I want to illustrate that the method is working, so I’ll graph a few tangent lines. By examining the derivative, I know the t value for which it is zero (t=1/2); from this I calculate the x and y values for which the function y(x) is a max/min. I also plot the tangent line at the arbitrary value t=2.

{x[1/2],y[1/2]}
p1=ParametricPlot[{x[t],y[t]},{t,.1,3}];
p2=Plot[y[2]+dydx[2]*(u-x[2]),{u,.1,3}];
p3=Plot[y[1/2]+dydx[1/2]*(u-x[1/2]),{u,.1,3}];
Show[p1,p2,p3]

#8

x[t_]:=1+Sqrt[t]
y[t_]:=Exp[t^2]
{x[1],y[1]}

dydx[t_]=Simplify[y'[t]/x'[t]]
dydx[1]
{x[1],y[1]}
p1=ParametricPlot[{x[t],y[t]},{t,.8,1.2}];
p2=Plot[y[1]+dydx[1]*(u-x[1]),{u,1,3}];
Show[p1,p2]

#43:

x[t_]:=t*Sin[t]
y[t_]:=t*Cos[t]
t0=0;
t1=1;
ParametricPlot[{x[t],y[t]},{t,t0,t1}]
Integrate[Sqrt[x'[t]^2+y'[t]^2],{t, t0,t1}]