Math 181 homework 3 – solutions
Problem 1 Many plant seeds are propagated by wind. Eg linden tree
seeds are even equipped with ’propellers’ to slow down the fall, and of course
any horizontal breeze will carry the seed away from the tree. Suppose a leaf
falls off a linden tree. Let D be the horizontal distance that the leaf has from
its starting point, and H the height of the leaf above ground, both at time
t. Assume that at all times,
2
D2 + 4 = 10(16 − H) 3
a) Find the equation relating D0 (t) and H 0 (t).
b) If H 0 (t) = −0.3 meters per minutes, find D0 (t) at the time when H = 7.
a) All we have to do is differentiate the given equation with respect to t,
using the Chain Rule.
2DD0 =
1
20
(16 − H)− 3 (−H 0 ).
3
2
b) From the original equation with H = 7, we find D2 + 4 = 10 · 9 3 , so
q
2
D = 10 · 9 3 − 4 ≈ 6.27.
Then use this and the given values for H 0 and H in the equation found in
part a), so at that particular moment of time
q
20 − 1
2
2 10 · 9 3 − 4D0 =
· 9 3 · 0.3.
3
We can solve this for D0 to get
1
0
D =p
9− 3
2
10 · 9 3 − 4
≈ 0.077.
Problem 2 Consider the function
f (x) = x2 e−4x
a) Find f 0 (x).
b) Find all critical numbers of f (x).
c) Find where f (x) is increasing and where it is decreasing.
d) Find an equation for the tangent line to f (x) at x = 1.
e) Sketch the graph of f (x) and the tangent line from d). Your x-axis should
include all critical numbers and a little bit more each way.
a) We don’t have to factor f 0 (x) but it helps for b)-c).
f 0 (x) = (−4x2 + 2x)e−4x
b) No points where f 0 (x) does not exist. So we only need to solve f 0 (x) = 0
which can only happen when
−4x2 + 2x = 0
therefore the only critical numbers are x = 0 and x = 0.5.
c) The sign patter of f 0 (x) is the same as that of −4x2 + 2x, namely − + −.
So the function f (x) is decreasing for x < 0 and x > 0.5, increasing for
0 < x < 0.5.
d) The slope is f 0 (1) = −2e−4 ≈ −0.039 and it passes through the point
(1, f (1)) = (1, e−4 ). So an equation for the tangent line is (any form is OK)
y = −2e−4 (x − 1) + e−4 = e−4 (−2(x − 1) + 1) = e−4 (−2x + 3)
≈ −0.039(x − 1) + 0.055.