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)
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!
I apologize, but as an AI text-based bot, I cannot view or access direct links or images. However, I can provide a verbal description of how the slope field and graph could look like.
For the slope field in question 1E, based on the given differential equation dy/dx = x(cube root (y)), you would have to select 9 points on the xy-plane and calculate the slope at each point using the given equation. Then, you can draw short line segments with the corresponding slopes at each of those points.
For question 2D, to graph the number of cougars as a function of time, you can set up a coordinate system with the number of cougars, C(t), on the y-axis and the time, t, on the x-axis. Start by plotting the given points (0, 1000) and (5, 2000) on the graph. Then, using the equation C(t) = 3500 - 2500 e^((-ln(5)+ln(3))/5 * t), you can plot additional points and connect them to create a smooth curve on the graph.
I hope this helps! Let me know if you have any further questions.
I'm sorry, but as an AI text-based bot, I am unable to see or interpret images. However, I can provide you with some general guidance on how to draw a graph of C(t) as a function of time.
To draw the graph, follow these steps:
1. Identify the key points:
- C(0) = 1000 (initial population at t=0)
- C(5) = 2000 (population at t=5)
- Limit as t tends to infinity: 3500 (maximum population)
2. Plot the key points on the graph.
- Plot (0, 1000) as the starting point.
- Plot (5, 2000) as the next point.
3. Determine the trend of the graph.
- Since the population is increasing at a rate directly proportional to 3500 - C(t), the graph should show an upward trend.
- The graph should approach 3500 as t tends to infinity.
4. Connect the plotted points with a smooth curve that represents the trend of the function.
- Start from the initial point (0, 1000) and draw a curve that passes through the plotted points.
- The curve should slope upwards and approach 3500 as t increases.
Please note that without the specific values for t, it's challenging to provide an accurate graph. I recommend referring to the provided axes and using the information given to plot the graph accordingly.