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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#22}

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4.2, \#32.
$f(x)=2x^3-2x^2+1$ has derivative $f'(x)=6x^2-4x$.  In particular,
$f'(1)=6-4=2$.  Thus the tangent line at
$(1,1)$ has slope 2.  This is the line $y-1=2(x-1)$ or $y=2x-1$.

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4.2, \#38.
The graph of the parabola $f(x)=x^2-2x+4$ suggests there are
two lines through the origin which are tangent to the graph.
One way to find these lines:  (i) Write the equation of
the tangent line at the point $(a,f(a))$.  (ii) Determine
if the tangent line goes through $(0,0)$.  Indeed,
$f'(x)=2x-2$, so that $f'(a)=2a-2$.  Thus the tangent line at
$(a,f(a))=(a,a^2-2a+4)$ has slope $2a-2$.
The equation of this line is
$$y-(a^2-2a+4)=(2a-2)(x-a).$$
The
origin $(0,0)$ is on the line if $-(a^2-2a+4)=(2a-2)(-a)$.
This reduces to $a^2-4=0$, or $a=\pm2$.
If $a=2$, the point is $(2,4)$, the slope is 2, and the tangent
line is $y=2x$.
If $a=-2$, the point is $(-2,12)$, the slope is $-6$, and the tangent
line is $y=-6x$.

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4.2, \#42.  $\displaystyle T=2\pi\sqrt{\ell\over g}={2\pi\over\sqrt g}
\ell^{-1/2}$ and $\displaystyle {dT\over d\ell}=
{\pi\over\sqrt g}\ell^{-1/2}$.  Thus $dT/d\ell$ is positive, which
implies the period increases as the length of the pendulum increases.

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4.3, \#1, 15.
$$\eqalign{
    y&=5t^2+4e^t,\cr y'&=10t+4e^t,\cr} \qquad
  \eqalign{
    f(x)&=x^3+3^x\cr f'(x)&=3x^2+(\ln 3)3^x\cr}$$

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