Calculus 1 Assignment 6
Alex Cowan
[email protected]
Due Thuesday, November 10th at 5 pm
1. Differentiate the identity xy = 8 with respect to t to obtain
x
dy
dx
+y
=0
dt
dt
The problem statement says that x = 4, y = 2, and
values and solving for dx
dt in the equation above gives
dy
dt
= −3, and asks to find
dx
dt .
Substituting these known
dx
x dy
4
=−
= − · (−3) = 6
dt
y dt
2
2.
a) 1
1
b) 21 (The derivative of arctan(x) is 1+x
2)
df
df dg
0
0
0
c) (f (g(t))) = f (g(t))g (t) and dt = dg dt
d
arctan(et ) at t = 0 is 1 · 21 . We
d) If we set g(t) = et , and f = arctan, then a), b), and c) together tell us that dt
π
1
0
start at arctan(e ) = 4 , and increase by an amount 0.1 at a rate of 2 , so we end up at π4 + 0.05.
3.
a) The tangent line of the function f at the point a is f (a) + f 0 (a)(x − a). In this case, f (a) = log(α · 0 + 1) = 0
α
and f 0 (a) = α·0+1
= α, so the linear approximation is y = αx.
b) Write f (x) = αx + error. The limit is
lim
x→0
error
αx + error
= α + lim
x→0
x
x
If the second limit is 0. The only way it is something other than 0 is if error looks something like βx, but then
that would be part of the linear approximation (since it’s a line). Hence error = o(x), and the limit on the right
is 0.
4.
√
√
a) The rate of change of x at x = 9 is 12 √19 = 16 . So 9.01 is going to be 3 + 0.01
6 = 3.0016666.... The actual
answer is 3.0016662....
b) We should still approximate by the tangent line at x = 9,√since that makes numbers easy. The rate of change
is still 16 , so our estimate is 3.166.... This is too high, since x is concave down. I would guess that it’s too high
by 0.01. The actual answer is 3.162..., so the estimate is a fair bit better than I thought it was!
1
5.
a) There are several ways to do this. One way is to square both sides to get
2x6 + 3x2 + ... = 2x6 + (O(x))2 + x3 O(x) = 2x6 + O(x2 ) + O(x4 ) = 2x6 + O(x4 )
Since 3x2 = O(x4 ) and every step is reversible, this proves the statement.
q
√
Another way to do it is to factor out the leading term from the square root to get 2x3 1 + 2x3 4 . Linear
√
√
√
approximation says that 1 + u = 1 + O(u), so in this case we get 2x3 + x3 O x14 = 2x3 + O x1 , which is
substantially better than what we need.
You could also use the definition of big-O notation, by showing
√
√
2x6 + 3x2 − 4 − 2x3
<∞
lim
x→∞
x
and this isn’t too hard, but I prefer the other methods, personally.
1
1
9
8
The analogous statement would be (3x9 + 4x8 − x2 − 100) 5 = 3 5 x 5 + O(x 5 ) (in both cases the approximation we’re making is that the square root/fifth root is linear).
b)
√
lim
x→∞
2x3 + O(x) √
O(x) √
=
2
+
lim
= 2+0
x→∞ x3
x3
c)
1
9
8
1
9
8
3 5 x 5 + O(x 5 )
= lim 3 5 x 5 −r + O(x 5 −r )
r
x→∞
x→∞
x
lim
If r > 95 , then limit goes to 0. If r < 95 , then the limit goes to infinity. If r =
secondary term doesn’t end up mattering.
2
9
5,
1
then the limit is 3 5 . The