MAT 137Y: Calculus with proofs Assignment 4
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
MAT 137Y: Calculus with proofs
Assignment 4
Due on Tuesday, Dec 20 by 11:59pm via GradeScope
1. A cone-shaped drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the angle θ such that the cup has the maximum capacity and
the maximum capacity of such a cup.
Some formulas may be useful in the question: area of a sector with angle θ and radius r is ; the volume of a cone with bottom radius r and height h is Tr2 h; the surface area of a cone without the bottom is Tr^h2 + r2 .
Solution: Step 1: Let us call the radius of the bottom of the cone r and the height of the cone is h. Then the surface of the cone without the bottom is Tr^h2 + r2 . We notice that the area of the blue part in the disk is equal to Tr^h2 + r2 . We now try to find the area of the blue part in the disk. In fact, it’s the disk area minus the area of the sector with angle θ . Therefore, we have
Tr ^h2 + r2 = TR2 - R2
Also we have
h2 + r2 = R2
And we want to maximize the volume of the cone.
V = Tr2 h
Step 2: Now we want to use (1) and (2) to rewrite the volume of cone as a function of θ . First, we can simplify the left hand side of (1) by (2)
Tr ^h2 + r2 = Tr^R2 = TrR
Thus, we have
θ
Now we can solve r as a function of θ
θ
2T
Second, we can use (2) to solve h as a function of r
Thus, we have
h = ^R2 - r2 = ′R2 - R2(1 - )2 = R ′ - = ^4πθ - θ 2
Finally, we can write the volume of the cone as a function of θ with θ e (0, 2π)
V (θ) = πR2(1 - )2 . ^4πθ - θ 2
R3 (2π - θ)2 ^4πθ - θ 2
=
6 4π2
= (2π - θ)2 ^4πθ - θ 2
Step 3: Finding critical points. Now this problem is a usual optimization question. We need to find the critical points of V and then check the critical points. First, we take derivative:
V\ (θ) = . [-2(2π - θ) ^4πθ - θ 2 + (2π - θ)2 . (4πθ - θ 2 )一1/2(4π - 2θ)]
R3 (2π - θ)(3θ2 - 12πθ + 4π2 )
= 24π2 . ^4πθ - θ 2
Note that ^4πθ - θ 2 0 and 2π - θ 0 when θ e (0, 2π). Now we can find the critical points by letting 3θ2 - 12πθ + 4π2 = 0. By using the quadratic formula, we have
12π 士 ^(12π)2 - 4 . 3 . 4π2 2
6 3
Since θ e (0, 2π), we have exactly one critical point
θ = 2π - ^6π
Step 4: Interpreting the result.
● When 0 < θ < 2π - ^6π , V\ (θ) > 0 and V is increasing. You can choose a sample point at and V\ ( ) > 0.
● When 2π - ^6π < θ < 2π , V\ (θ) < 0 and V is decreasing. You can choose a sample point at and V\ ( ) < 0.
This means V has a maximum at θ = 2π - ^6π (both a local maximum and a global maximum).
And
2 2^3πR3
3 27
2. A water tank has the shape of a horizontal cylinder with radius R = 1 m and length 2 m. If water is being pumped into the tank at a rate of m3 per minute, find the rate at which the water level is rising when the water is m deep.
Step 1: Let h(t) be the height of the water filled in the tank at time t. Let V (t) be the volume of the water filled in the tank at time t. We have known = m3 per minute. We want to find when h = m.
Step 2: Finding a relationship between h and V . Here, we only consider the case when 0 < h < 1. See the picture below and we will find the area of the blue segment. We call it Aˆ . Therefore, the volume of the water will be the area of the blue segment times the length of the tank, i.e. V = 2Aˆ .
From the picture, we can observe Aˆ = area of the sector OAB - area of the triangle △OAB .
To find the area of the triangle OAB, we can find PB = ^1 - (1 - h)2 and AB = 2PB = 2^1 - (1 - h)2 . The area of △OAB = . (1 - h) . 2 ^1 - (1 - h)2 .
To find the area of the sector OAB, we need to find the angle ZAOB first. Note that cos(ZPOB) = 1 - h. Thus, we have ZAOB = 2ZPOB = 2 arccos(1 - h). Therefore, the area of the sector OAB = . ZAOB . 12 = arccos(1 - h).
Finally, we have
Aˆ = arccos(1 - h) - (1 - h) ^1 - (1 - h)2
and
V = 2Aˆ = 2[arccos(1 - h) - (1 - h) ^1 - (1 - h)2].
Step 3: finding the unknown rate by taking derivatives with respect to time t. Now we can take
derivatives on both sides of (6) with respect to t and we can get
= 2 ┌ +′1 - (1 - h)2 - ┐ = 2 ┌ +′1 - (1 - h)2 ┐
= 2 ┌ ′ 1 - (1 - h)2 +′1 - (1 - h)2 ┐
= 4 ′ 1 - (1 - h)2
Now we can plug h = and = in (10). We have
= 4 ′ 1 -
Finally, we have = 81^3 or m per minute.
The water level is rising at a rate of m per minute when the water is meter deep.
Remark: we can also use the same idea to find the rate of change for h when h > . For example, we can find the rate of change for h when h = . Now the volume of the water in terms of h is different but the rate of the change of h when h = is the same as the rate of the change of h when h = .
3. We can use Rolle’s Theorem to prove a generalized Mean Value Theorem.
Theorem 1. 夕桂t a < b | 夕桂t f αn占 g )桂 two 大unétions w在ié在 αr桂 éontinuous on [a, b] αn占 占i务桂r桂ntiα))桂 on (a, b)| 4ssum桂 g(b) g(a) αn占 g\ (x) 0 大or αnz x e (a, b)| T在桂n t在桂r桂 桂αists ξ e (a, b) sué在 t在αt
f (b) - f (a) f\ (ξ)
=
g(b) - g(a) g\ (ξ)
(a) We can find a specific g(x) in the theorem to get the standard Mean Value Theorem. What is
g(x)? No need to justify your answer to this part. g(x) = x
(b) Prove the theorem 1.
Hint: Define a new function F (x) = f (x) 一(一)g(x) or H(x) = g(x)[f (b) - f (a)] - f (x)[g(b) -
g(a)].
Method 1:
Solution: Define H(x) = g(x)[f (b) - f (a)] - f (x)[g(b) - g(a)] on [a, b].
Since f and g be two functions which are continuous on [a, b] and differentiable on (a, b), we can conclude H(x) is continuous on [a, b] and differentiable on (a, b).
Note
H(a) = g(a)[f (b) - f (a)] - f (a)[g(b) - g(a)] = g(a)f (b) - f (a)g(b)
and
H(b) = g(b)[f (b) - f (a)] - f (b)[g(b) - g(a)] = -g(b)f (a) + f (b)g(a)
Thus, we have H(a) = H(b). Applying the Rolle’s Theorem for H(x) on [a, b], we can conclude there exists ξ e (a, b) such that H\ (ξ) = 0. Therefore, we get
f (b) - f (a) f\ (ξ)
g(b) - g(a) g\ (ξ)
Note that 一(一) = is well-defined since g(b) g(a) =告 g(b) - g(a) 0 and g\ (x) 0 for
any x e (a, b) =告 g\ (ξ) 0.
This completes the proof.
Method 2:
Solution: Define F (x) = f (x) 一(一)g(x) on [a, b]. Since g(b) g(a), this function is well-
defined.
Since f and g be two functions which are continuous on [a, b] and differentiable on (a, b), we can conclude F (x) is continuous on [a, b] and differentiable on (a, b).
Note
f (b) - f (a) f (a)g(b) - f (b)g(a)
g(b) - g(a) g(b) - g(a)
and
f (b) - f (a) -f (b)g(a) + f (a)g(b)
g(b) - g(a) g(b) - g(a)
Thus, we have F (a) = F (b). Applying the Rolle’s Theorem for F (x) on [a, b], we can conclude there exists ξ e (a, b) such that F\ (ξ) = 0. Therefore, we get
f (b) - f (a) f (b) - f (a) f\ (ξ)
g(b) - g(a) g(b) - g(a) g\ (ξ)
Note that 一(一) = is well-defined since g(b) g(a) =告 g(b) - g(a) 0 and g\ (x) 0 for any x e (a, b) =告 g\ (ξ) 0.
4. Now let’s use Question 3 to prove one case of the L’H pital’s Rule.
Theorem 2. 夕桂t a < b | 夕桂t f αn占 g )桂 two 大unétions w在ié在 αr桂 占桂/n桂占 αn占 占i务桂r桂ntiα))桂 on (a, b)| 4ssum桂
● g αn占 g\ αr桂 n桂v桂r 0 on (a, b))
● lim f (x) = lim g(x) = 0) x→a+ x→a+
f\ (x)
x→a+ g(x) x→a+ g\ (x) |
To prove this theorem, we need to use this following lemma:
夕桂mmα Let a < b. Let H be a function defined on (a, b). For each x e (a, b), let cx be a real number
such that a < cx < b. If lim H(x) exists, then lim H(cx) = lim H(x).
x→a+ x→a+ x→a+
(a) Assume that f and g are also defined at a and f (a) = g(a) = 0. With this extra hypothesis, prove
Theorem 2.
βroo大| – Let x e (a, b).
– Since f and g are differentiable on (a, b) and (a, x) c (a, b), we can say f and g are differen- tiable on (a, x).
– Since we have lim f (x) = lim g(x) = 0 and f (a) = g(a) = 0. We can conclude that f x→a+ x→a+
and g are continuous at a. Also, since f and g are differentiable at x, we have f and g are continuous at x. Therefore, f and g are continuous on [a, x].
– Note that g(x) 0 = g(a) and g\ (y) 0 for any y e (a, x).
– Now we can apply Theorem 1 in Q3 on [a, x]. We conclude that there exists ξ e (a, x) such
that
f (x) - f (a) f\ (ξ)
=
g(x) - g(a) g\ (ξ)
So we have
f (x) f (x) - f (a) f\ (ξ)
x→a+ g(x) x→a+ g(x) - g(a) x→a+ g\ (ξ)
since f (a) = g(a) = 0. Use the lemma given with H(x) = |
f\ (x) g\ (x) |
and cx = ξ . Thus, we have |
f\ (ξ) f\ (x)
x→a+ g\ (ξ) x→a+ g\ (x)
Therefore, we have proved
f (x) f\ (x)
x→a+ g(x) x→a+ g\ (x)
(b) Now prove Theorem 2 without any extra hypotheses.
五int| Define a new function such that = f for all x e (a, b) and (a) = 0. Do the same
βroo大| – Let x e (a, b).
– Define such that = f for all x e (a, b) and (a) = 0.
x→a+ (x) x→a+ \ (x)
x→a+ g(x) x→a+ g\ (x)
5. Suppose that f is a function with domain R. Is each of the following claims true or false? If it is true, prove it. If it is false, provide a counter example and prove that it satisfies the required conditions. Hint: you may want to use L’Hoˆpital’s Rule in this problem.
f (x)
x→& x
Final Answer This claim is TRUE FALSE.
Solution:
βroof| Since f has a slant asymptote as x - o, we can find y = mx + b with m, b e R and such that
lim [f (x) - (mx + b)] = 0
x→&
In this case, lim = 0 because the numerator has limit 0 and the denominator
has limit o. Then
f (x) f (x) - (mx + b) + (mx + b)
x→& x x→& x
= xl ┌ + m + ┐
= 0 + m + 0 = m
f (x)
x→& x
x→& x
Final Answer This claim is TRUE FALSE.
There are many counter examples. Here is one: f (x) = cosx. We will prove lim = 0 and
cos has no slant asymptote as x - o.
cos x
x→& x
By the Squeeze Theorem, since
-1 sin x 1
< <
x x x
for all x > 0 and
lim -1 = lim 1 = 0,
we can conclude lim = 0.
● Claim 2: cos does not have a slant asymptote as x - o. We want to prove that for every m, b e R
lim [cos x - (mx + b)] 0
x→&
Notice that
xl ┌ - m - ┐ = 0 - m + 0 = -m
Therefore
xl[cos x - (mx + b)] = xlx . ┌ - m - ┐ = 士o
where the limit is o if m < 0, -o if m > 0 and the limit does not exist if m = 0. In all three cases the limit does not exist and not equal to 0.
Remark: There are many other counterexamples: f (x) = sinx, f (x) = x + cos x, f (x) = ^x, or f (x) = lnx.
(c) Let a e R. Suppose f is differentiable on R. If lim f′ (x) = o, then lim = o.
βroof| Since lim f′ (x) = o, we have
x→a
lim = lim f′ (x) = o
Hence we can apply L’Hoˆpital’s Rule:
lim = lim = lim f′ (x) = o
2023-01-13