Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/167

Fourth step.	${\frac {dx}{dt}}$	= 2 ft. per second
	$x=6\$.	${\frac {dy}{dt}}$	= ?
	$y={\frac {x^{2}}{6}}=6$.	${\frac {ds}{dt}}$	= ?
Fifth step. Substituting back in (B),
	${\frac {dy}{dt}}={\frac {6}{3}}\times 2=4$ ft. per second. Ans.

From the first result we note that at the point P(6, 6) the ordinate changes twice as rapidly as the abscissa.

If we consider the point P'(-6, 6) instead, the result is ${\tfrac {dy}{dt}}=-4$ ft. per second, the minus sign indicating that the ordinate is decreasing as the abscissa increases.

3. A circular plate of metal expands by heat so that its radius increases uniformly at the rate of .01 inch per second At. what rate is the surface increasing when the radius is two inches?

Circular plate of metal.

Solution. Let x = radius and y = area of plate. Then
	$\ y$	$=\pi x^{2}\$.
(C)	${\frac {dy}{dt}}$	$=2\pi x{\frac {dx}{dt}}$,

That is; at any instant the area of the plate is increasing in square inches $2\pi x$ times as fast as the radius is increasing in linear inches.

	$\ x$	$=2,{\frac {dx}{dt}}=.01,{\frac {dy}{dt}}=?$
Substituting in (C),
	${\frac {dy}{dt}}$	$=2\pi \times 2\times .01=.04\pi$ sq. in. per sec. Ans.

4. An arc light is hung 12 ft. directly above a straight horizontal walk on which a boy 5 ft. in height is walking. How fast is the boy's shadow lengthening when he is walking away from the light at the rate of 168 ft. per minute?

Arc light over a straight walk.

Solution. Let x = distance of boy from a point directly under light L, and y = length of boy's shadow. From the figure,
	$y:y+x::5:12\$,
or	$\ y$	$={\frac {5}{7}}x$.
Differentiating,	${\frac {dy}{dt}}$	$={\frac {5}{7}}{\frac {dx}{dt}}$;

i.e. the shadow is lengthening ${\tfrac {5}{7}}$ as fast as the boy is walking, or 120 ft. per minute.

5. In a parabola $y^{2}=12x$, if x increases uniformly at the rate of 2 in. per second, at what rate is y increasing when x = 3 in. ? Ans. 2 in. per sec.