2.3 Homework
Andy Long
Spring, 2016

#56, 60, 92, p. 136

#56: remember that the normal line is perpendicular to the tangent line, and they both pass through the same point (in this case, that’s (1,9)).

f[x_]:=(1+2x)^2
f'[x]
x0=1
f[1]
m=f'[x0] (* here's the slope of the tangent line *)
-1/m (* and the slope of the normal line is the negative multiplicative inverse *)
tangent[x_]:=m(x-x0)+f[1]
normal[x_]:=(-1/m)(x-x0)+f[1]
tangent[x]
normal[x]
Plot[{f[x],tangent[x],normal[x]},{x,x0-0.1,x0+0.1},AspectRatio->Automatic]

#60: it’s easy to calculate derivatives with Mathematica:

g[r_]:=r^(1/2)+r^(1/3)
g'[r]
g''[r]
Simplify[g'[r]]
Simplify[g''[r]]

#92: What it means to say that f(20)=10000, and f’(20)=-350, is that f is on the decrease from 10000 bolts of cloth at a price of p=20; in fact, for a one-unit change in p, you’d expect about a -350 change in the number of bolts of cloth produced.

r[p_]:=p*f[p]
(* To carry out this problem, I can actually model f by a function that satisfies the conditions of the problem. I made a linear function that meets the function value f(20)=10000 and and the slope f'(20)=-350. This would be the tangent line to any function f that actually pertains to this problem. *)
f[p_]:=(-350)*(p-20)+10000
r'[20]
(* I get a value of 3000: what does it mean? It means that if we increase the price, although we'll reduce the quantity of cloth, the increased price will mean greater overall revenue (roughly 3000 for a one-dollar increase in price). *)