Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

Sec. 9] Miscellaneous Exercises on First-Order Differential Equations 343
2893. Find the curve, of which the segment of the tangent
contained between the coordinate axes is divided into half by
the parabola if =--2x.
2894. Find the curve whose normal at any point of it is
equal to the distance of this point from the origin.
2895*. The area bounded by a curve, the coordinate axes,
and the ordinate of some point of the curve is equal to the
length of the corresponding arc of the curve. Find the equation
of this curve if it is known that the latter passes through the
point (0, 1).
2896. Find the curve for which the area of a triangle formed
by the x-axis, a tangent, and the radius vector of the point of
tangency is constant and 2
equal to a .
2897. Find the curve if we know that the mid-point of the
segment cut off on the x-axis by a tangent and a normal to the
curve is a constant point (a, 0).
When forming first-order differential equations, particularly in phvsical
problems, it is frequently advisable to apply the so-called method of differentials,
which consists in the fact that approximate relationships between
infinitesimal h.crements of the desired quantities (these relationships are
accurate to infinitesimals of higher order) are replaced by the corresponding
relationships between their differentials. This does not affect the result.
Problem. A tank contains 100 litres of an aqueous solution containing
10 kg of salt. Water is entering the tank at the rate of 3 litres per minute,
and the mixture is flowing out at 2 litres per minute. The concentration is
maintained uniform by stirring. How much salt will the tank contain at the
end of one hour?
Solution. The concentration c of a substance is the quantity of it in
unit volume. If the concentration is uniform, then the quantity of substance
in volume V is cV.
Let the quantity of salt in the tank at the end of t minutes be x kg.
The quantity of solution in the tank at that instant will be 100 + / litres,
and, consequently, the concentration c= QQ
kg per litre.
During time dt, 2dt litres of the solution flows out of the tank (the
solution contains 2cdt kg of salt). Therefore, a change of dx in the quantity
of salt in the tank is given by the relationship
This is the sought -for differential equation. Separating variables and integrating,
we obtain
ln* = 21n(100+0 + lnC
or
C
*
'
1
(100-M)
The constant
C= 100,000. At
C is
the
found from
expiration
the
of
fact
one
that xvh^n f = 0, \ 10,
hour, the tank will
that is,
contain
x
100,000
-
ft
=^ 3.9
. ..
f
kilograms of
u
salt.