Math 1A with Prof. Bamler, DIS 112 and 113midterm 2 review

April 12th, 2018

1 Find the following limits.

1. limx→1−2x2+x−1x2+x−2

2. limθ→0sin2 θ

2θ

3. limx→−∞√

x2 + 3x + 1− x

4. limx→∞x2+3

x3−3x2+1

5. limx→∞e√

ln x

x

1. If x → 1−, the denominator x2 + x− 2 = (x− 1)(x + 2) ≈ 0 and is negative. The numerator2x2 + x− 1 ≈ 2. Therefore this limit is −∞.

2. Write it as lim sin θ2 ·

sin θθ = lim sin θ

2 · limsin θ

θ = 0 · 1 = 0.

3. limx→−∞√

x2 + 3x + 1 = +∞ and limx→−∞(−x) = +∞ so the limit is +∞.

4. The highest power of x on the denominator is large than the highest power of x on thenumerator for this rational function. Therefore the limit is 0.

5. Write x = eln x to get

limx→∞

e√

ln x

eln x = limx→∞

e√

ln x−ln x

= limx→∞

e√

ln x(1−√

ln x).

Let y =√

ln x(1 −√

ln x). Then as x → ∞, we have y → −∞ since√

ln x → ∞ and1−√

ln x → −∞. Therefore the limit is limy→−∞ ey = 0.�

2 Find the equation of the tangent line to the curve x2 − xy− y2 = 1 at the point (2, 1).

Implicitly differentiate with respect to x to get 2x − (y + xy′)− 2yy′ = 0. At the point (2, 1), wehave 2 · 2− (1 + 2y′)− 2y′ = 0 so 4− 1− 4y′ = 0 so 4y′ = 3 so y′ = 3/4. The tangent line is theline which has slope 3/4 and passes through (2, 1), which is given by y− 1 = (3/4)(x− 2). �

3 Differentiate.

1. f (x) =√

x−1x5+1

2. f (x) = (tan x)1/x

Use logarithmic differentiation for both.

1. We have ln f (x) = 12 ln( x−1

x5+1 ) =12 (ln(x− 1)− ln(x5 + 1)). Differentiate to get f ′(x)/ f (x) =

12 (

1x−1 −

5x4

x5+1 ) so f ′(x) =√

x−1x5+1 ·

12 (

1x−1 −

5x4

x5+1 ).

2. We have ln f (x) = (1/x) ln tan x. Therefore f ′(x)/ f (x) = (−1/x2) ln tan x + (1/x) sec2 xtan x .

Therefore f ′(x) = (tan x)1/x((−1/x2) ln tan x + (1/x) sec2 xtan x ).

�

4 Water is being poured at a rate of 40cm3/s into a funnel in the shape of an upside-down conewhich has base diameter 5cm and height 10cm. Water exits the bottom of the funnel at a rate of20cm3/s. How fast is the height of the water changing when the water is overflowing out of thecone? (The volume of a cone with base radius r and height h is V = 1

3 πr2h.)

The volume of the water is increasing at a total rate of 40 − 20cm3/s which is 20cm3/s. Let Vbe volume of the water and h the height of the cone and r the radius of the base of the cone. Bysimilar triangles, we must have r = h/4. The volume of the water is then

V =13

πr2h =13

π

(h4

)2

h =1

3 · 16πh3.

Differentiate with respect to time t to get

dVdt

=1

16πh2 dh

dt.

Since dV/dt = 20cm3/s and we want dh/dt at h = 10, which is when the water is overflowing,we get

20 =116

π · 100dhdt

(1)

dhdt

=20 · 16100π

(2)

=165π

. (3)

�

5 Sketch the graph of the function

f (x) =ex

x2

on its domain by finding the following information:

1. domain

2. horizontal and vertical asymptotes

3. intervals of increase/decrease

4. absolute and local maxes and mins (for those that exist)

5. intervals of concavity and inflection points

6. sketch of the graph (label intercepts, extrema, inflection points)

A formula for your convenience: f ′′(x) = (x2−4x+6)ex

x4 .

1. The domain of the function is x such that x 6= 0.

2. Let’s find the horizontal asymptotes. First:

limx→∞

ex

x2 = limx→∞

ex

2xby L’Hospital’s rule (4)

= limx→∞

ex

2by L’Hospital’s rule again (5)

= ∞. (6)

Second:

limx→−∞

ex

x2 = limx→∞

1x2ex (7)

= 0. (8)

The only candidate for a vertical asymptote is the line x = 0. We check:

limx→0+

ex

x2 =≈ 1≈ 0+

= ∞, (9)

limx→0−

ex

x2 =≈ 1≈ 0+

= ∞. (10)

3. To find intervals of increase/decrease, we differentiate.

f ′(x) =x2ex − 2xex

x4 =(x− 2)ex

x3 . (11)

This has one root, at x = 2. If we look at the sign of f ′ at the intervals (−∞, 0), (0, 2), and(2, ∞), we get:

f ′(x) =

+ if x ∈ (−∞, 0) ∪ (2, ∞)

0 if x = 2− if x ∈ (0, 2).

(12)

This gives us the intervals of increase/decrease.

4. The intervals of increase/decrease tells us by the first derivative test that f (2) = e2

4 is a localminimum. There is no local maximum. We know that limx→∞

ex

x2 = ∞ so there is no absolutemaximum. We also know that ex

x2 is always > 0 but limx→−∞ex

x2 = 0 so there is also noabsolute minimum.

5. We differentiate once more to get

f ′′(x) =x3(x− 1)ex − (x− 2)ex(3x2)

x6 (13)

=(x4 − x3 − 3x3 + 6x2)ex

x6 (14)

=(x2 − 4x + 6)ex

x4 . (15)

Look at the roots of x2 − 4x + 6 = 0. See that x2 − 4x + 6 = (x − 2)2 + 2 ≥ 2 > 0. Sinceex is also always positive, f ′′(x) > 0 for all x. So f is concave up everywhere. There are noinflection points.

6. You can graph it using your computer to see what it looks like. You can put all the previousinformation together to get something close to the actual graph.

�

6 A right circular cylinder is inscribed in a sphere of radius R. Find the largest possible volumeof such a cylinder.

The volume of a right circular cylinder is πr2h where h is the height and r is the radius of thecylinder (notice that this is different from R). Looking at the cylinder in the sphere from the side,we measure the angle θ created by drawing a line from the center of the sphere to the corner of thecylinder. Then the height of the cylinder is given by h = 2R sin θ and the radius by r = R cos θ. Sothe volume of the cylinder is V = 2πR3 cos2 θ sin θ. Differentiate in θ to obtain

V ′ = 2πR3(−2 sin3 θ + cos3 θ).

If V ′ = 0, then cos3 θ = 2 sin3 θ. Therefore

tan θ =sin θ

cos θ=

121/3 .

Therefore θ = arctan 121/3 is the volume maximizing angle. Now recall that the volume of the

cylinder is V = 2πR3 cos2 θ sin θ. Find cos θ and sin θ and plug it back in. �