MATH 31A: MIDTERM 2 REVIEW
JOE HUGHES
1. The Chain rule
1. Compute
d x+sin x
dx tan(2x) .
Solution: Let f (x) = x + sin x and g(x) = tan(2x). Then f 0 (x) = 1 + cos x, and g 0 (x) =
2 sec2 (2x) by the chain rule. Therefore by the quotient rule,
(1 + cos x)(tan(2x)) − (x + sin x)(2 sec2 (2x))
d x + sin x
=
dx tan(2x)
tan2 (2x)
sin x
We can simplify this expression by recalling that tan x = cos
x and sec x =
2
multiplying the numerator and denominator by cos (2x) yields
1
cos x .
So
(1 + cos x)(tan(2x)) − (x + sin x)(2 sec2 (2x))
(1 + cos x) sin(2x) cos(2x) − 2(x + sin x)
=
2
tan (2x)
sin2 (2x)
q
2. If f (x) =
x+
p
√
x3 + x5 , find f 0 (1).
Solution: Using the chain rule several times, we get
r
q
√
d
1
1
5x4 2
3
5
q
p
x+ x + x =
√ · 3x + √ 5
p
√ · 1+
dx
3
2 x
2
x
+
x5
2 x + x3 + x5
At this point, it makes more sense to plug in x = 1 than to try to simplify the above
expression. So
√
1 + 411
3 + 52 1
2
0
= p
f (1) = p
√ · 1+ √
√
2
2
2 1+ 2
2 1+ 2
2. Implicit Differentiation
1. Find the tangent line of the curve defined by the equation x2 y − x + tan(y − 1) = 0 at
the point (1, 1).
Solution: Using implicit differentiation, we obtain
1
2
JOE HUGHES
2xy + x2
dy
dy
− 1 + sec2 (y − 1)
=0
dx
dx
dy
You could solve for dx
at this point, but if you only need to find it at a single point, it’s
easier to plug in x = 1, y = 1 first. So
2+
and
dy
dx |(1,1)
dy
dy
−1+
=0
dx
dx
= − 12 . Therefore the tangent line has equation
1
y − 1 = − (x − 1)
2
2. Show that no point on the curve x2 − y 2 = 1 has a tangent line with slope 1.
Solution: Using implicit differentiation shows that
2x − 2y
dy
=0
dx
dy
so that dx
= xy . If the tangent line to the curve at a point P = (x, y) had slope one, this
would imply that x = y. But then
x2 − y 2 = 0 6= 1
so P is not on the curve.
If you remember your conic sections, this result makes sense. x2 − y 2 = 1 is a hyperbola,
and its asymptotes are the lines y = ±x. So it is plausible at least that no point on the
curve should have a tangent line with slope 1.
3. Related Rates
1. Ship A is traveling west toward a lighthouse at a speed of 15 kilometers per hour, while
ship B is traveling north away from the lighthouse at a speed of 10 kilometers per hour.
Write x for the distance between ship A and the lighthouse, and y for the distance between
ship B and the lighthouse.
a. Find the rate of change of the distance between the two ships when x = 4 km and y = 3
km.
Solution: Write d(t) for the distance between the ships. Then by the Pythagorean theorem,
d(t)2 = x(t)2 + y(t)2
Differentiating this equation with respect to time gives
2d(t)d0 (t) = 2x(t)x0 (t) + 2y(t)y 0 (t)
MATH 31A: MIDTERM 2 REVIEW
3
Now solve for d0 (t):
d0 (t) =
x(t)x0 (t) + y(t)y 0 (t)
d(t)
Finally, we√need to plug in the values of x, y, etc. At the given time, x = 4 and y = 3,
hence d = 9 + 16 = 5. Since ship A is approaching the lighthouse, x(t) is decreasing, so
x0 (t) = −15, while y is increasing, so y 0 (t) = 10. Therefore
d0 (t) =
4(−15) + 3(10)
30
= − = −6
5
5
in units of km/hr.
b. Let θ be the horizontal angle in the right triangle formed by the two ships and the
lighthouse. Find the rate of change of θ when x = 4 km and y = 3 km.
Solution: tan θ = xy , so differentiating with respect to time gives
sec2 (θ)θ0 (t) =
y 0 (t)x(t) − y(t)x0 (t)
x(t)2
Therefore
y 0 (t)x(t) − y(t)x0 (t)
cos2 (θ)
x(t)2
We saw in part (a) that if x = 4 and y = 3 then d = 5, hence cos θ = 45 . Thus
85
17
(10)(4) − 3(−15) 4 2 40 + 45 16
·
=
·
=
=
θ0 (t) =
16
5
16
25
25
5
in units of radians per hour.
θ0 (t) =
2. A 15 foot long ladder is leaning against a building. Let x(t) denote the distance of the
end of the ladder touching the ground from the base of the building, and let y(t) denote
the distance of the end of the ladder touching the building from the base. Suppose that
x0 (t) = 21 feet per second.
a. Find y 0 (t) when x = 9.
Solution: The ladder forms a right triangle with the building and the ground. The hypotenuse (i.e. the ladder) has constant length, hence
x(t)2 + y(t)2 = 152
Differentiating gives
2x(t)x0 (t) + 2y(t)y 0 (t) = 0
and after solving for y 0 ,
y 0 (t) = −
If x = 9 then
y=
√
x(t)x0 (t)
y(t)
225 − 81 =
√
144 = 12
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JOE HUGHES
so
9 · 21
3
=−
y (t) = −
12
8
0
in units of feet per second.
b. Find the rate of change of the area of the triangle when x = 9.
Solution: The area of the triangle is A(t) = 12 x(t)y(t). Differentiating using the product
rule, we obtain
1
1
A0 (t) = x0 (t)y(t) + x(t)y 0 (t)
2
2
From part (a), we know that if x = 9 then y = 12 and y 0 (t) = − 83 . Therefore
1
3
27
21
1 1
· · 12 + · 9 · (− ) = 3 −
=
2 2
2
8
16
16
in units of square feet per second.
A0 (t) =
4. Linear Approximation
1. Estimate
√
4
80 using linear approximation.
Solution: The key to any linear approximation problem is finding the function f and the
1
point a to use. In this case we should take f (x) = x 4 , and a = 81.
1
f (a) = 81 4 = 3, and f 0 (x) =
tangent line is therefore
1 − 34
4x
=
1
3
4x 4
, so f 0 (a) =
1
4·27
=
1
108 .
The equation of the
x−a
108
1
and plugging in x = 80 shows that f (80) ≈ L(80) = 3 − 108 = 323
108 .
L(x) = f (a) + f 0 (a)(x − a) = 3 +
If you’re short on time during the exam, just leave the answer as 3 −
1
108 .
5. Critical Points and the first derivative test
2
1. Find the critical points of f (x) = x 3 , and determine whether they are local maxima or
minima (or neither).
1
1
Solution: f 0 (x) = 23 x− 3 for x 6= 0, and f 0 (0) does not exist. Since 23 x− 3 is non-zero for all
x 6= 0, this implies that the only critical point is c = 0.
To determine whether 0 is a minimum or a maximum, use the first derivative test: if x < 0
1
1
then x− 3 < 0, while if x > 0 then x− 3 > 0. Therefore c = 0 is a minimum.
2. Find the critical points of f (x) = sin2 x, and determine whether they are local maxima
or minima (or neither).
MATH 31A: MIDTERM 2 REVIEW
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Solution: f 0 (x) = 2 sin x cos x, so the critical points of f are points where either sin x = 0
or cos x = 0. sin(x) = 0 for x = 0, ±π, ±2π, . . . and cos(x) = 0 for x = ± π2 , ± 3π
2 ,....
Using a trig identity, we can write f 0 (x) = sin(2x), so f 00 (x) = 2 cos(2x). Thus f 00 (nπ) =
2 cos(2nπ) = 2 > 0 for any integer n, so the points c = 0, ±π, ±2π, . . . are local minima.
On the other hand, if n is odd then f 00 ( nπ
2 ) = 2 cos(nπ) = −2 < 0, so the points c =
± π2 , ± 3π
,
.
.
.
are
local
maxima.
2
3. Let f (x) = x3 + bx2 + cx + d be a cubic polynomial. Show that f has at most 2 local
extrema.
Solution: f 0 (x) = 3x2 + 2bx + c is a quadratic polynomial, hence has at most two roots.
Therefore f has at most two critical points, so at most two local extrema.
6. The mean value theorem
1. Let f (x) = sin(tan(x)). Prove that there exists some real number c such that f 0 (c) =
4 sin(1)
.
π
Solution: This problem is a bit tricky because you have to find the right interval on which
to apply the Mean Value Theorem. But the fraction π4 suggests that the interval should
have length π4 .
So let’s try the interval [0, π4 ]. Then
f ( π4 ) − f (0)
sin(tan π4 ) − sin(tan(0))
4 sin(1)
=
=
π
π
π
4 −0
4
hence the Mean Value Theorem implies that there is a point c in (0, π4 ) with f 0 (c) =
4 sin(1)
.
π
2. Let f (x) be continuous on [0, 1] and differentiable on (0, 1), with f (0) = 0 and f (1) = 1.
Show that there is a point c in (0, 1) such that f 0 (c) = 2c.
Solution: Let g(x) = f (x) − x2 . Then g(0) = 0 and g(1) = f (1) − 1 = 1 − 1 = 0. By the
Mean Value Theorem applied to g, there is a point c in (0, 1) such that
g 0 (c) =
g(1) − g(0)
=0
1−0
Therefore
0 = g 0 (c) = f 0 (c) − 2c
which shows that f 0 (c) = 2c.
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JOE HUGHES
7. Concavity and the second derivative test
1. Let f (x) = x2x+1 . Find the inflection points of f , the intervals on which f (x) is concave
up and concave down, and use the second derivative test to find the local maxima and
minima of f .
2
2
=
Solution: The derivative is f 0 (x) = x(x+1−2x
2 +1)2
exists. So the critical points are c = ±1
1−x2
,
(1+x2 )2
which is zero at x = ±1 and always
The second derivative is
(−2x)(1 + x2 )2 − 2(1 − x2 )(1 + x2 )(2x)
(−2x)(1 + x2 ) − 4x(1 − x2 )
=
(1 + x2 )4
(1 + x2 )3
2(x3 − 3x)
−2x3 − 2x − 4x + 4x3
=
(1 + x2 )3
(1 + x2 )3
√
√
√
which is zero at x =√0, ± 3. f is concave down
on
(−∞,
3),
concave
up
on
(−
3, 0),
√
concave down on (0, 3), and concave up on ( 3, ∞).
=
Finally, f 00 (1) < 0, so c = 1 is a max, while f 00 (−1) > 0, so c = −1 is a min.
8. Optimization
1. A tank with a rectangular base and rectangular sides is to be open at the top. It is to
be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building
the tank costs 10 dollars per square meter for the base and 5 dollars per square meter for
the sides, what are the dimensions of the least expensive tank, and what is its cost?
Solution: Let l, w, h denote the length, width, and height of the tank. Then the cost of the
tank is
C = 10lw + 2 · 5lh + 2 · 5wh = 10(lw + lh + wh)
This our objective function, i.e. the function we want to minimize. Of course, to apply the
techniques we’ve learned, we need this function to depend on only one variable.
The first step is to recall that w = 4, so the objective function becomes
C = 10(4l + lh + 4h)
Next, the volume of the tank is 36 cubic meters, so
36 = lwh = 4lh
Therefore l =
9
h,
so
36
+ 9 + 4h)
h
Now take the derivative and set it equal to zero:
360
0 = C 0 (h) = − 2 + 40
h
C(h) = 10(
MATH 31A: MIDTERM 2 REVIEW
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Solving for h gives
360
=9
40
so h = 3 (the negative square root can be ignored, since the tank can’t have negative
height).
h2 =
Now take the second derivative and evaluate it at h = 3:
720
C 00 (3) =
>0
27
so the second derivative test guarantees that this is a minimum.
Thus
l=
9
9
= =3
h
3
and the cost of the tank is
10(
(in dollars).
36
+ 9 + 4h) = 10(12 + 9 + 12) = 330
h