Solutions For Calculus Quiz #2

Solutions For Calculus Quiz #2

1. (a) To solve the equation, we separate variables first and then integrate. We find Z

dy = Z √

3x + 1 dx.

If we set

u = 3x + 1 with du = 3 dx, then

y(x) = Z √

u1 3 du

= 1 3· 2

3(u^3/2) + C

= 2

9(u)^3/2+ C

= 2

9(3x + 1)^3/2+ C.

Using the given condition y(0) = 1 allows us to determine C:

1 = 2 9+ C

C = 7

9. The solution is therefore

y(x) = 7 9 +2

9(3x + 1)^3/2.

(b) To solve the equation, we separate variables first and then integrate. We find Z dy

y + 1 = Z

e^−x dx.

Carrying out the integration on both sides, we obtain ln |y + 1| = −e^−x+ C1. Solving for y, we find

|y + 1| = exp(−e^−x+ C1) y + 1 = ± exp(−e^−x+ C1) y + 1 = ±e^C1 exp(−e^−x).

Setting C = ±e^C1, we obtain

y = −1 + C exp(−e^−x).

Using the given condition y(0) = 2 allows us to determine C:

2 = −1 + C exp(−1) C = 3e.

The solution is therefore

y = −1 + (3e) exp(−e^−x) y = −1 + 3 exp(1 − e^−x).

(c) Separation of variables yields

Z dy

y(3 − y) = Z

2 dx.

We use the partial fraction method to integrate the left-hand side.

1

y(3 − y) = A

y + B 3 − y

= A(3 − y) + By y(3 − y)

= (B − A)y + 3A y(3 − y) Comparing the last term to the integrand, we find

B − A = 0 and 3A = 1, and thus

A = 1

3 and B = 1 3.

Using the partial fraction decomposition, we must integrate 1

3 Z

(1 y + 1

3 − y) dy = Z

2 dx which yields

1

3(ln |y| − ln |3 − y|) = 2x + C¹. Simplifying this results in

ln | y

3 − y| = 6x + 3C¹

| y

3 − y| = e^3C1e^6x y

3 − y = ±e^3C1e^6x y

3 − y = Ce^6x. Using the given condition y(1) = 5, we find

C = −5 2e⁻⁶. The solution is therefore

y

3 − y = −5

2e⁻⁶e^6x.

If we want to write the solution in the form y = f (x), we must solve y.

We find

y = −5

2e⁻⁶e^6x(3 − y) y(1 −5

2e⁻⁶e^6x) = −15 2 e⁻⁶e^6x y = −¹⁵₂e⁻⁶e^6x 1 −⁵2e⁻⁶e^6x

y = 3

1 −²₅e^−6(x−1).

2. (a) In this case L∞ = 123, and the equation becomes dL

dt = k(123 − L(t)) with L(0) = 1.

To solve the equation, we separate variables and integrate. This yields

Z dL

123 − L(t) = Z

k dt.

Hence,

− ln |123 − L(t)| = kt + C1. After multiplying this equation by -1 and exponentiating,

|123 − L(t)| = e^−C1e^−kt or

123 − L(t) = Ce^−kt with C = ±e^−C1.

Since L(0) = 1, we find

C = 122.

The solution is then given by

123 − L(t) = 122e^−kt or

L(t) = 123 − 122e^−kt. Now substituting L(27) = ¹²³₂ = 61.5, we get

61.5 = 123 − 122e^−27k e^−27k = 61.5

122 and taking ln on both sides we have

27k = ln 122 61.5 k ≈ 0.0254.

(b) The length of the fish after 10 months is

L(10) = 123 − 122e^{−0.0254×10}

≈ 28.37 (inches).

(c) Suppose at time t the fish reaches 90% of the asymptotic length.

To find t, we solve the equation

0.9 × 123 = 123 − 122e^−0.0254t e^−0.0254t = 12.3

122

t = 1

0.0254ln 122 12.3 t ≈ 90.33 (months).

3. (a) We define g(x, y) = 1 x + 1

y − 1, x 6= 0, y 6= 0.

Then the constraint is of the form g(x, y) = 0.

Using the method of Lagrange multipliers, we are looking for (x, y) and λ so that

▽f (x, y) = λ▽g(x, y) and g(x, y) = 0.

Since

▽f (x, y) =1 1



and ▽g(x, y) =−_x¹²

−_y¹²



this translates into the set of equations 1 = − λ

x² , 1 = −λ

y² ,and 1 x+ 1

y = 1.

We can eliminate λ from the first two equations. We find x² = y² or x = ±y.

Combining this with the constraint equation, we obtain the candidate (x, y) = (2, 2).

To classify (x, y) = (2, 2), we write f (x, y) as a function of one variable x, f (x, y) = x + y

= x + x x − 1

= x² x − 1. Let

g(x) = x² x − 1, and compute

g^′(x) = 2x(x − 1) − x² (x − 1)²

= x²− 2x (x − 1)²

= x(x − 2) (x − 1)²

g^′′(x) = (x − 1)²2(x − 1) − (x²− 2x)2(x − 1) (x − 1)⁴

= 2(x − 1) (x − 1)⁴. we find that

g^′(2) = 0 , and g^′′(2) = > 0, therefore (x, y) = (2, 2) is a local minimum.

(b) There is no global extrema because

x→1lim⁺g(x) = lim

x→1⁺

x²

x − 1 = ∞, and

x→1lim⁺g(x) = lim

x→1⁻

x²

x − 1 = −∞.

4. We find

▽Y (N, P ) =(1 − N)P e^{−(N +P )} (1 − P )Ne^{−(N +P )}



=0 0



when

(1 − N)P e^{−(N +P )} = 0 and (1 − P )Ne^{−(N +P )} = 0.

This implies

(N, P ) = (0, 0) or (N, P ) = (1, 1).

Since Y (N, P ) is differentiable on the domain,

there is only two critical point, namely, (0,0) and (1,1).

To classify the critical point, we compute Hess Y (N, P ) =

 −P e^{−(N +P )}− (1 − N)P e^{−(N +P )} (1 − N)e^{−(N +P )}− (1 − N)P e^{−(N +P )} (1 − P )e^{−(N +P )}− (1 − P )Ne^{−(N +P )} −Ne^{−(N +P )}− (1 − P )Ne^{−(N +P )}

 . Evaluating this at the critical point (0,0), we find

Hess Y (0, 0) = 0 1 1 0

 . Consider

1 −10 1 1 0

  1

−1



= 1 −1−1 1



= −2

1 10 1 1 0

 1 1



= 1 11 1



= 2, we find that the curvature along two respective direction are different, therefore (0,0) is a saddle point.

Evaluating Hess Y (N, P ) at the critical point (1,1), we find Hess Y (1, 1) = −e⁻² 0

0 −e⁻²

 .

Since det Hess Y (1, 1) > 0, and tr Hess Y (1, 1) < 0, then (1,1) maximizes Y . 5. (a) Since p3 = 1 − p¹− p², we have

H(p1, p2) = −p¹ln p1− p²ln p2 − p³ln p3

= −p¹ln p1− p²ln p2 − (1 − p¹− p²) ln(1 − p¹− p²).

As the function ln x is only defined for x > 0, we must have p1 > 0, p2 > 0, and p1+ p2 < 1 which construct the domain of H(p1, p2).

(b) We find

▽H(p1, p2) =− ln p1− 1 + ln(1 − p¹− p²) + 1

− ln p²− 1 + ln(1 − p¹− p²) + 1



=

"

ln(^1−p_p¹₁^−p2) ln(^1−p_p¹₂^−p2)

#

=0 0



when

1 − p¹− p² = p1 and 1 − p¹− p² = p2. This implies

p1 = 1/3 and p2 = 1/3.

Since H(p₁, p₂) is differentiable on the domain, there is only one critical point, namely, (1/3,1/3).

To classify the critical point, we compute Hess H(p1, p2) =

" _p1

1−p1−p2[^{−p1−(1−p}_p2 ^1−p2)

1 ] −_1−p^p1¹−p2

−_1−p^p1²−p2

p2

1−p1−p2[^{−p2−(1−p}_p2 ^1−p2)

2 ]

# .

Evaluating this at the critical point (1/3,1/3), we find Hess H(1/3, 1/3) =−6 −1

−1 −6

 . Since det Hess H(1/3, 1/3) > 0 and tr Hess H(1/3, 1/3) < 0, then H has a local maximum at (1/3,1/3).

But (1/3,1/3) is the only one critical point of H,

therefore H also attains its absolute maximum at (p1, p2) = (1/3, 1/3).

6. (a) To verify that

c(x, t) = 1

√4πDtexp[− x² 4Dt]

is the solution to the partial differential equation, we compute

∂c(x, t)

∂t = [ 1

√4πDt]^′exp(− x²

4Dt) + 1

√4πDt[exp(− x² 4Dt)]^′

= 1

√4πDt(−1

2)t^−3/2exp(− x²

4Dt) + 1

√4πDt(−x²

4D)(−t⁻²) exp(− x² 4Dt),

∂c(x, t)

∂x = 1

√4πDt(− 2x

4Dt) exp(− x²

4Dt), and

∂²c(x, t)

∂x² = 1

√4πDt(− 2x

4Dt)^′exp(− x²

4Dt) + 1

√4πDt(− 2x

4Dt)[exp(− x² 4Dt)]^′

= 1

√4πDt(− 2

4Dt) exp(− x²

4Dt) + 1

√4πDt(− 2x

4Dt)(− 2x

4Dt) exp(− x² 4Dt)

= 1

D[ 1

√4πDt(−1

2)t^−3/2exp(− x²

4Dt) + 1

√4πDt(−x²

4D)(−t⁻²) exp(− x² 4Dt)]

= 1

D

∂c(x, t)

∂t .

Therefore c(x, t) is the solution to the partial differential equation

∂c(x, t)

∂t = D∂²c(x, t)

∂x² .

(b)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

0 0.5 1 1.5 2 2.5 3

t=1 t=0.1

t=0.01