Question

1. The expression dy/dx = x(cube root (y)) gives the slope at any point on the graph of the function f(x) where f(2) = 8.

A. Write the equation of the tangent line to f(x) at point (2, 8).
y'=x∛y
The tangent line at (2,8) is thus
y-8 = 4(x-2)

B. Write an expression for f(x) in terms of x.
dy/dx = x y^(1/3)
y^(-1/3) dy = x dx
(3/2)y^(2/3) = (1/2)x^2 + c
when x = 2 , y = 8
(3/2)(4) = (1/2)(4) + c
c = 6 - 2 = 4
(3/2) y^(2/3) = (1/2) x^2 + 4
3 y^(2/3) = x^2 + 8
y^(2/3) = (1/3) x^2 + 8/3

C. What is the domain of f(x)?
x^2+8 >= 0, or all reals.

D. What is the minimum value of f(x)?
for minimum value, y'=0, or where x=0 or y=0. Since y is never zero, the minimum is at x=0.

E. Using the axes provided, sketch a slope field for the given differential equation at the nine points indicated.
(gyazo.com/d1944f7a206262301c82db884f090464)
please help with this one!

2. Let C(t) be the number of cougars on an island at time t years (where t > 0). The number of cougars is increasing at a rate directly proportional to 3500 - C(t). Also, C(0) = 1000, and C(5) = 2000.

A. Find C(t) as a function of t only.
C'(t) = k ( 3500 - C(t))
C'(t) = 3500k - kC(t)
C'(t) + kC(t) = 3500k

Multiply by e^(kt):
e^(kt)(C'(t) + kC(t) ) = 3500k e^(kt)
(e^(kt)C(t))' = 3500k e^(kt)

Integrate both sides:
e^(kt)C(t) = 3500e^(kt) + A
C(t) = 3500 + Ae^(-kt)

Set t = 0:
1000 = 3500 + A
A = -2500

Set t = 5:
2000 = 3500 - 2500 e^(-5k)
2500e^(-5k) = 1500
e^(-5k)= .6
-5k = ln .6
k = -(ln .6)/5
C(t) = 3500 - 2500 e^((ln .6)/5 * t)

B. Calculate C(10).
(is there an easier way to calculate this?)
dC / dt = k ∙ ( 3500 - C(t) )
dC = k ∙ ( 3500 - C(t) ) ∙ dt
dC / ( 3500 - C(t) ) = k ∙ dt
∫ dC / ( 3500 - C(t) ) = k ∙ ∫ dt
− ln | 3500 - C(t) | = k⋅t + h
h = constant of integration
Multiply both sides by - 1
ln | 3500 - C(t) | = - k⋅t - h
mark:
- h = m
ln | 3500 - C(t) | = - k⋅t + m
3500 - C(t) = e⁻ᵏᵗ⋅eᵐ
mark:
eᵐ = A
3500 - C(t) = e⁻ᵏᵗ⋅A
- C(t) = A⋅ e⁻ᵏᵗ - 3500
Multiply both sides by - 1
C(t) = 3500 - A⋅ e⁻ᵏᵗ
Now:
t = 0
C(0) = 1000
1000 = 3500 - A⋅ e⁰
1000 = 3500 - A⋅ 1
1000 = 3500 - A
A = 3500 - 1000
A = 2500

t = 5
C(5) = 2000
2000 = 3500 - A⋅ e⁻ ⁵ᵏ
2000 = 3500 - 2500⋅e⁻ ⁵ᵏ
(2000 - 3500) / -2500 = e⁻ ⁵ᵏ
-1500 / -2500 = e⁻ ⁵ᵏ
-3 ∙ 500/-5 ∙ 500 = e⁻ ⁵ᵏ
-3/-5 = e⁻ ⁵ᵏ
3/5 = e⁻ ⁵ᵏ
e⁻ ⁵ᵏ = 3/5
ln (e⁻ ⁵ᵏ) = ln(3/5)
-5k = ln(3/5)
k = ln (3/5) / -5
k = [ ln(3) - ln(5) ] / -5
k = [ ln(5) - ln(3) ] / 5
k = 0.102165125
C(t) = 3500 - 2500⋅e⁻⁰⋅¹⁰²¹⁶⁵¹²⁵ ᵗ
C(10) = 3500 - 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ = 2600

C. Find the limit as t tends to infinity of C(t) , and explain its meaning.
lim ( 3500 - 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ ᵗ) = 3500 -t->∞
lim (- 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ ᵗ) = t->∞
3500 - 2500 e⁻∞ = 3500 - 2500⋅0 = 3500 - 0 = 3500
The number of cougars on island can't be greater of 3500.

D. On the axes provided, draw a graph showing the number of cougars as a function of time:
answer: (gyazo.com/131f514a5369d94ee8e26225070c8e56)

Answer 1

1E just compute y' at each indicated point, just as you did for (2,8).
Draw a small line segment with that slope at each point

#2 Solving the 1st-order DE, you get
C(x) = 3500 + c*e^(-kx)
C(0) = 1000 means C(x) = 3500 - 2500e^(-kx)
C(5) = 2000 means C(x) = 3500 - 2500e^(-0.102x)

so, it's easy to figure C(10) = 3500 - 2500e^-1.02 = 2598.51
Not sure why you went to all that trouble when you had the formula!

Answer 2

I'm having a hard time with 1E... do you think you could sketch it out... I'm not really sure how to visualize it. Sorry about this and thanks for the help!