Selected Answers to Practice Midterm # 2 (The actual midterm will be shorter!) Math 220, Fall 2010
1. Find the first and second derivatives of the following functions. Simplify where appropriate. (a) y = x2 e5x . y ′ = 2xe5x + 5x2 e5x = x(2 + 5x)e5x . y ′′ = 2e5x + 20xe5x + 25x2 e5x = (25x2 + 20x + 2)e5x .
(b) y = x3 cos(x2 + 4) . y ′ = 3x2 cos(x2 + 4) − 2x4 sin(x2 + 4) .
y ′′ = 6x cos(x2 + 4) − 14x3 sin(x2 + 4) − 4x5 cos(x2 + 4) . (c)
√ x2 + 1 Simplify! y= x x 1 (x2 + 1)−1/2 2x − (x2 + 1)1/2 1 y′ = 2 . =− √ 2 x x2 x2 + 1 y ′′ =
3x2 + 2 . x3 (x2 + 1)3/2
2. Differentiate the following functions. Simplify where appropriate. (a) y = arccot(3 arccos(4x)) .
y′ =
12 √ . (1 + 9(arccos(4x))2 ) 1 − 16x2
(b) y = ln(1 + xex ) . (1 + x)ex . 1 + xex
y′ =
(c) y = (csc(2x))(cot(3x))(sec(4x)) . y ′ = − 2 csc(2x) cot(2x) cot(3x) sec(4x) − 3 csc(2x) csc2 (3x) sec(4x) + 4 csc(2x) cot(3x) sec(4x) tan(4x) . (d) x2 + 4 √ y= x2 + 1 y′ =
(e)
(f)
x(x2 − 2) . (x2 + 1)3/2
√ x2 + 3 y= 2 x +2 y′ = −
y =e
Simplify!
x(x2 + 4) √ . (x2 + 2)2 x2 + 3
√ e2x x2 + 1 y= x ′
Simplify!
2x
Simplify!
2x3 + 2x − 1 √ x2 x2 + 1
.
(g) y=
′
y =x
2
x3 cos(3x) . 1 + x2 e5x
3(1 + x2 e5x )(cos(3x) − x sin(3x)) − x2 e5x cos(3x)(2 + 5x) (1 + x2 e5x )2
(Not much you can do with that...) (h) y = sin(5 sin(6 cos(7x))) .
y ′ = −210 cos(5 sin(6 cos(7x))) · cos(6 cos(7x)) · sin(7x) . (i) y = sin(x2 sin(x3 cos(4x))) . This one will not fit on my computer screen without some creativity. So ... Let A(x) := sin(x3 cos(4x)) and B(x) := cos(x3 cos(4x)) , then y ′ = cos(x2 A(x))· 2xA(x) + x2 B(x) · 3x2 cos(4x) − 4x3 sin(4x) and Kudos to you if you simplified this and got:
y ′ = x cos(x2 A(x)) · 2A(x) + x3 B(x) · (3 cos(4x) − 4x sin(4x)) .
Also, notice that it is easy to mess up the parenthesis! What I have above is correct and can also be written: y ′ = x cos(x2 A(x))· 2A(x) + x3 B(x) · (3 cos(4x) − 4x sin(4x)) . But it is not equal to: x cos(x2 A(x)) · 2A(x) + x3 B(x) · (3 cos(4x) − 4x sin(4x)) .
(j) y = e( 2+x cos x ) . 1+x sin x
′
y =
x2 + x sin x + 2x cos x + 2 sin x − cos x (2 + x cos x)2
e( 2+x cos x ) . 1+x sin x
(k) 1 + tan(4x) . y = arctan 2 + sec(e5x ) Again this one will not fit without renaming some things, so ...
let 1
A(x) := 1+
1+tan(4x) 2+sec(e5x )
(2 + sec(e5x ))2 , (2 + sec(e5x ))2 + (1 + tan(4x))2
2 =
B(x) := 4[2 + sec(e5x )] sec2 (4x) , and C(x) := 5e5x (1 + tan(4x)) sec(e5x ) tan(e5x ) . Then with these definitions, we have: ′
y = A(x)
B(x) − C(x) (2 + sec(e5x ))2
.
Except for my simplification of A(x), I can see nothing particularly worth simplifying. (l) sin y= Let A(x) :=
e3x (x2 +1)17
1 − cos 5x
.
e3x (3x2 − 34x + 3) e3x d = , dx (x2 + 1)17 (x2 + 1)18
then y′ =
[1 − cos(5x)] cos
e3x (x2 +1)17
A(x) − 5 sin(5x) sin
(1 − cos(5x))2
e3x (x2 +1)17
.
3. Derive the formulas for the derivatives of tan x, cot x, sec x, csc x, arcsin x, and arccos x. The derivation of the formula for the derivative of tan x can be found in section 3.3 at the bottom of page 192. The derivation of the formula for the derivative of cot x is very similar. The derivation of the formula for the derivative of arcsin x, can be found in section 3.5 at the top of page 212. The derivation of the formula for the derivative of arccos x is very similar. Since the derivation of the formula for the derivative of sec x is very similar to the derivation of the formula for csc x, I will simply do the derivation for cosecant: d 1 d csc x = dx dx sin x (sin x) · 0 − 1 · (cos x) = (sin x)2 − cos x = (sin x)2 1 cos x · =− sin x sin x = − cot x csc x .
by the quotient rule
dy 4. Find dx for the following implicitly defined functions at the point in question.
(a) x2 + y 2 = 25 at the points (4, 3) and (4, −3) . Differentiation w.r.t. x gives us: 2x + 2y
dy =0 dx
which leads to
dy x =− . dx y
Now simply plug in the points (4, 3) and then (4, −3) to get − 43 and 34 respectively.
(b) x2 + y 2 = 169 at the points (5, 12) and (5, −12) . Here again you will get dy x =− . dx y 5 5 and 12 respectively. Now plug in (5, 12) and (5, −12) to get − 12 (You can check your work for the last two problems by solving for y and differentiating in the “old-fashioned” way. Be careful about the sign of the square root.) √ √ Well by solving for y you get either y = 25 − x2 or y = − 25 − x2 for part (a), and then y ′ = −x(25 − x2 )−1/2 or y ′ = x(25 − x2 )−1/2 respectively. Of course now there is only “x” to √ plug in, and in both (b) you get y = 169 − x2 or √ cases x = 4. For part ′ 2 y = − 169 − x , and then y = −x(169 − x2 )−1/2 or y ′ = x(169 − x2 )−1/2 . For each of the points here, x = 5.
(c) x3 y 2 − x2 y 3 + xy = 6 at the point (2, 1) . Differentiation w.r.t. x gives us: 3x2 y 2 + 2x3 y
dy dy dy − 2xy 3 − 3x2 y 2 +y+x =0, dx dx dx
and then we can solve for
dy dx
to get:
2xy 3 − 3x2 y 2 − y dy = 3 , dx 2x y − 3x2 y 2 + x which means (plugging in the point (2, 1)) dy 4 − 12 − 1 9 3 = =− =− . dx 16 − 12 + 2 6 2
(d) yexy = 2e2
at the point (1, 2) .
Differentiation w.r.t. x gives us: dy dy xy xy =0, y+x e + ye dx dx and then we can solve for
dy dx
to get:
dy y2 =− , dx 1 + xy which means (plugging in the point (1, 2)) 4 4 dy =− =− . dx 1+2 3 (e) y sin
πx 4
+
πy = 2 at the point (−2, 4) . 6
Differentiation w.r.t. x gives us: πx πy πx πy π π dy dy =0, sin + + + + y cos dx 4 6 4 6 4 6 dx and then we can solve for dy = dx sin
dy dx
to get:
− πy cos 4 πy πx + 6 + 4
πx + πy 4 6 πy πx cos 6 4
+
which means (plugging in the point (−2, 4)) √ √ −π2 3 dy −3π 3 √ = √ . = π 3 1 dx 3 + 2π 3 + 2 3
πy 6
,
dy 5. Find dx using logarithmic differentiation. (i.e. Take the logarithm and dy .) then differentiate, and then solve for dx
(a) y = xx . dy = xx (1 + ln x) . dx (b) y = (2x)(sin x) . sin x dy (sin x) cos x ln(2x) + = (2x) . dx x (c) y = (3x + 4)(5x+6) . dy 3(5x + 6) (5x+6) . 5 ln(3x + 4) + = (3x + 4) dx 3x + 4 (d) sin7 (5x)(ln(1 + x6 ))1/3 y= . (x2 + 5)9 e(x4 +1) sin7 (5x)(ln(1 + x6 ))1/3 dy = · A(x) , dx (x2 + 5)9 e(x4 +1) where A(x) :=
18x 2x5 − 2 − 4x3 35 cot(5x) + 6 6 (1 + x ) ln(1 + x ) x + 5
.
6. For this problem you should know that the volume of a ball of radius r is 34 πr3 and its surface area is 4πr2 . You will also need: 2.12 = 4.41, 2.52 = 6.25, 2.13 = 9.261, 2.53 = 15.625. (Leave π as π everywhere it occurs!) Give the units of your final answers!
(a) Find the average rate of change of the area of a square with respect to its side length, s, as s changes from i. 2 in to 3 in A(s) = s2 so the average r.o.c. is 32 − 22 = 5 square inches per inch = 5 in . 3−2
ii. 2 in to 2.5 in
iii. 2 in to 2.1 in
2.25 2.52 − 22 = = 4.5 in . 2.5 − 2 .5
.41 2.12 − 22 = = 4.1 in . 2.1 − 2 .1 (b) Find the instantaneous rate of change of area of the square with respect to its side length when s = 2 in. (Note that this is half of the perimeter of the square.) A′ (s) = 2s so A′ (2) = 4 square inches per inch or 4 in. (c) Find the average rate of change of the volume of a ball with respect to its radius, r, as r changes from i. 2 in to 3 in V (r) = 34 πr3 so the average r.o.c. is 4 π(27 3
− 8) 76π 76π 2 = cubic inches per inch = in . 3−2 3 3
ii. 2 in to 2.5 in
4 π(15.625 3
− 8) 61π 2 = in . 2.5 − 2 3
iii. 2 in to 2.1 in 4 π(9.261 3
− 8) 4π12.61 50.44π 2 = = in . 2.1 − 2 3 3
(d) Find the instantaneous rate of change of volume of the ball with respect to radius when r = 2 in. (Note that this is the surface area of the sphere.) V ′ (r) = 4πr2 so V ′ (2) = 16π cubic inches per inch or 16π in2 .
(e) Find the average rate of change of the volume of a cube with respect to its side length, s, as s changes from i. 2 in to 3 in V (s) = s3 so the average r.o.c. is 27 − 8 = 19 cubic inches per inch = 19in2 . 3−2
ii. 2 in to 2.5 in
iii. 2 in to 2.1 in
15.625 − 8 = 15.25in2 . 2.5 − 2
9.261 − 8 = 12.61in2 . 2.1 − 2 (f) Find the instantaneous rate of change of volume of the cube with respect to its side length when s = 2 in. (Note that this is half of the surface area of the cube.) V ′ (s) = 3s2 so V ′ (2) = 12 cubic inches per inch or 12 in2 . 7. A stone is dropped onto a lake, creating a circular ripple that travels outward at a speed of 50 in/sec. Find the rate at which the area within the circular ripple is increasing after 1 sec, 3 sec, and 5 sec. Give the units of your final answers! dr dA = 2πr . dt dt dr = 50 so the radii are at the times in question 50, 150, and 250 inches dt (by using distance equals rate times time), and then the answers are (in order) 5000π, 15,000π, and 25,000π square inches per second. A = πr2
so
8. A tank which holds 1000 gallons of water drains according to the formula: 2 t V = 1000 1 − 0 ≤ t ≤ 20 20 where V is volume of water remaining in the tank at time t (given in minutes). What is the rate at which water is flowing out of the tank after 5 min, 10 min, 15 min, and 20 min? Give the units of your final answers!
dV 1 t t − = 100 = 1000 · 2 1 − −1 dt 20 20 20 is the r.o.c. of volume of water in the tank. On the other hand, since we want the rate at which water is flowing out of the tank, this number is off by a minus sign. The function we want is therefore: t , F (t) = 100 1 − 20 and plugging in gives us 75, 50, 25, and 0 gallons per minute. 9. A machine inflates a spherical balloon at a rate of 5in3 /sec. How fast is the radius increasing at the moment that the radius is 4in? How fast is the surface area increasing at the same moment? Give the units of your final answers! 4 dV dr V = πr3 so = 4πr2 . 3 dt dt dV dr 5 Since dt = 5 and r = 4 plugging in gives us dt = 64π inches per second as the answer to the first question. Next: S = 4πr2 Now we plug in r = 4 and second.
dr dt
so =
dr dS = 8πr . dt dt 5 64π
to get
dS dt
=
5 2
square inches per
10. A 15ft ladder is leaning against a wall. The bottom of the ladder slides away at a constant rate of 2ft/sec. Find the rate at which the top of the ladder slides down the wall at the instant when the bottom of the ladder is 9ft from the wall. Give the units of your final answers! According to your model, what happens to the speed of the top of the ladder as the bottom of the ladder gets close to being (but less than) 15ft away from the wall? What can you conclude? Let x denote the distance from the base of the ladder to the wall, and let y denote the height on the wall where the top of the ladder is in
contact. Then x2 + y 2 = 152 = 225 so 2x Solving for
dy dt
and plugging in
dx dt
dx dy + 2y =0. dt dt
= 2, gives
dy −2x = . dt y In particular, at the moment x = 9, by the Pythagorean Theorem we get y = 12, and so by plugging in we get dy = − 32 which means that dt the ladder is sliding down the wall at a rate of 1.5 feet per second. For the next question, since y approaches 0 as x approaches 15, if the model were to hold until the top of the ladder hit the ground, then the velocity would go to infinity. Certainly, there is no way that this would happen and so there must be something wrong with our model. In fact, what is wrong is simply that when the ladder is falling fast enough it will separate from the wall rather than increase its speed to maintain contact with the wall. 11. A spotlight lying on the ground shines on a wall 25ft away. A 10ft tall giraffe walks from the wall toward the spotlight at a speed of 3ft/sec. At what rate is the height of its shadow on the wall increasing at the instant when it is 15ft away from the light? Give the units of your final answers! Let y denote the height of the shadow, and let x denote the distance from the spotlight to the spot on the ground directly below the head of the giraffe. By using similar triangles we get the equation: 10 y = = 10x−1 25 x
so
1 dy dx = −10x−2 . 25 dt dt
We have dx = −3, and x = 15 and so by plugging in we get dt feet per second.
dy dt
=
10 3
12. A hose fills a conical tank (a right circular cone with the “tip” of the cone at the bottom) with the radioactive liquid “New Jerseyium” at a rate of 10ft3 /min. The tank has height h = 50ft and radius r = 40ft at the top. How fast is the fluid rising at the moment that the height is
5ft? Give the units of your final answers! (Note that the volume, V, of a right circular cone with height, h, and radius, r, is given by the formula: 1 V = πr2 h . 3 Note also that h and r are related by similar triangles in a problem of this type.) Let h denote the height of the fluid and let r denote the radius of the 40 fluid at the top. Using similar triangles we see hr = 50 = 45 and so we have 2 dV 16πh2 dh 4h 1 16πh3 and so = . V = π h= 3 5 75 dt 25 dt Plugging in we get
dh dt
=
10 16π
=
5 8π
feet per minute.
13. A plane leaves Phoenix at 3 PM travelling North at 400miles/hour. At 4 PM another plane leaves Phoenix travelling East at 300miles/hour. At what rate is the distance between these planes increasing at 5 PM? What if the first plane started at 2 PM instead of 3 PM? Give the units of your final answers! Making obvious definitions for x and y, and letting d denote the distance between the airplanes we have p y = 400t , x = 300(t − 1) , and d = x2 + y 2 where t = 0 denotes 3 PM. Then dx 300(t − 1) · 300 + 400t · 400 dy ′ 2 2 −1/2 . x +y d (t) = (x + y ) = p dt dt (300(t − 1))2 + (400t)2 Now plug in t = 2 to get the correct number of miles per hour.
For the next part of the problem it makes more sense to measure time from 2 PM. With this change we have y = 400t and x = 300(t − 2) and plug in t = 3. 14. Particle on a curve problems...
(a) If y = x2 + 4x + 2, and
dx dt
= 5, then what is
dy dx = (2x + 4) dt dt √ (b) If y = 2 x, and
dx dt
dx dy = x−1/2 dt dt (c) If y = e4x , and
dx dt
dx dy = 4e4x dt dt √ (d) If y = x2 + 1, and x → ∞?
dx dt
dy dt
so
when x = 9?
dy 4 = . dt 3
so
= 3, then what is
when x = 1?
dy = 30 . dt
so
= 4, then what is
dy dt
dy dt
when x = 2?
dy = 12e8 . dt
= 17, then to what does
dy dt
converge as
dx x x dy dy . =√ so = 17 lim √ x→∞ dt dt x2 + 1 dt x2 + 1 The limit is slightly tricky. By observing that √ p √ √ x = x2 ≤ x2 + 1 ≤ x2 + 2x + 1 = (x + 1)2 = x + 1, one can use the squeeze theorem to show that the limit is 1. Therefore the answer is 17. 15. The element New Jerseyium has a half-life of 10 days. A sample originally has a mass of 500 grams. (a) Find a formula for the mass remaining after t days. ( 10t ) −t ln 2 1 = 500e( 10 ) . A(t) = 500 2
(b) When is the mass 10 grams? 10 = 500e(
−t ln 2 10
)
⇒
t=
10 ln 50 . ln 2
16. A common inhabitant of the bowels of government buildings is the bacterium “Bureaucratiazoa.” Sadly, this bacteria doubles its population every 5 years. A sample found in a building in the state of Taxachusetts has an initial population of 50. (a) Find a formula for the population in t years. A(t) = 50 (2)( 5 ) = 50e( t
t ln 2 5
).
(b) When does the population reach 500? 500 = 50 (2)( 5 ) t
⇒
t=
5 ln 10 . ln 2
17. On planet X if a stone is thrown upward with an initial velocity of v0 (given in feet per second), then the height of the stone at time t is given by h(t) = v0 t − 2t2 . (I am assuming that the stone is being thrown from height zero.) (a) If v0 = 12 feet per second, then how high will the stone get? h(t) = 12t − 2t2 , so h′ (t) = 12 − 4t. Setting h′ (t) = 0 we get t = 3. Then h(3) = 18 feet. (b) What will the velocity be when the stone first passes through the height of 16 feet? Setting h(t) = 16 we get the equation 2t2 − 12t + 16 = 0 which leads to (t − 2)(t − 4) = 0. h′ (2) = 4 feet per second. (c) What will the velocity be the second time that the stone has height 16? h′ (4) = −4 feet per second, or 4 feet per second toward planet X.
