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)

2 answers

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!
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!
Similar Questions
  1. Which expression simplifies to 87–√3?(1 point) Responses −67–√3−27–√3 negative 6 cube root of 7 minus 2 cube
    1. answers icon 1 answer
    1. answers icon 1 answer
  2. Which expression simplifies to 87–√3?(1 point) Responses −67–√3−27–√3 negative 6 cube root of 7 minus 2 cube
    1. answers icon 3 answers
  3. Which expression simplifies to 87–√3?(1 point) Responses 256−−√3−47–√3 2 cube root of 56 minus 4 cube root of 7
    1. answers icon 1 answer
more similar questions