Math 181 - homework 2 - solutions
.
Problem 1 Suppose the concentration of a pollutant in a pond is given as
a function of time t by
0.1t + 3
p(t) = 2
2t − t + 2
a) Find the derivative of p(t) using the Quick Rules.
b) Find the instantaneous rate of change of p(t) at t = 1 and t = 2.
c) Given that p(2) = 0.4, approximate the concentration at time t = 2.1
using the equation for the tangent line to the graph of p(t) at t = 2.
d) Find the interval(s) where p(t) is decreasing, and the interval(s) where it
is increasing.
e) When is the concentration of the pollutant maximal?
Solution. a) Use the Quotient Rule!
0.1(2t2 − t + 2) − (0.1t + 3)(4t − 1)
(2t2 − t + 2)2
2
−0.2t − 12t + 3.2
−0.2(t2 + 60t − 16)
=
=
(2t2 − t + 2)2
(2t2 − t + 2)2
p0 (t) =
(1)
b) As easy as substituting t = 1 and t = 2 into the above formula,
p0 (1) =
−9
= −1,
9
p0 (2) = −
21.6
= −0.3375
64
c) The equation for the tangent line is
y = −0.3375(t − 2) + 0.4
and if we plug in t = 2.1, we get y = 0.36625.
d) This means find intervals where p0 (x) > 0 (for increasing) and intervals
where p0 (x) is negative (for decreasing). So we look for singular points (there
are none, because the denominator is never zero here). Then we look for
stationary points, which means solving
t2 + 60t − 16 = 0
which has solutions
√
602 + 64
.
t1/2 =
2
The sign pattern of the derivative is − + −, with the boundaries between
intervals being the two t-values above. To see this, we could use our work
in part b): p0 (t) < 0 for all t > 0.27. Similarly, substituting t = 0 shows
p0 (t) > 0 for all t in (−60.27, 0.27) and p0 (t) < 0 for t < −60.27. So the
function p(t) is decreasing in (−∞, −60.27) and (0.27, ∞) and increasing in
(−60.27, 0.27).
It’s also OK although not specificallly stated if we think of this formula as
only valid for t > 0 and adjust the answer to decreasing in (0.27, ∞) and
increasing in (0, 0.27).
e) From the work in part d), it follows that p(t) has its (only) maximum at
t = 0.27.
−60 ±