Multiple Choice
Question 1
Which situation is modeled by dy/dt = k (a constant)?
A
a quantity changing at a constant rate
B
exponential growth
C
logistic growth
D
radioactive decay
Solution: A constant derivative means the quantity changes at a constant rate, giving a linear function y = kt + C.
Question 2
Solve dy/dx = e^x with y(0) = 3.
A
y = e^x + 2
B
y = e^x + 3
C
y = e^x
D
y = e^x - 2
Solution: Antidifferentiate: y = e^x + C. Using y(0) = 3: 1 + C = 3, so C = 2 and y = e^x + 2.
Question 3
If dy/dt = 0.05y and y(0) = 200, what is y(t)?
A
200 e^(0.05t)
B
200 + 0.05t
C
200 e^(5t)
D
e^(0.05t) + 200
Solution: Exponential growth solution: y = y0 e^(0.05t) = 200 e^(0.05t).
Question 4
Which function is a solution of dy/dx = -2y?
A
y = e^(-2x)
B
y = e^(2x)
C
y = -2x
D
y = x^(-2)
Solution: y = e^(-2x) gives dy/dx = -2 e^(-2x) = -2y, satisfying the equation.
Question 5
Solve dy/dx = (2x)/(y).
A
y^2 = 2x^2 + C
B
y^2 = x^2 + C
C
y = x^2 + C
D
y = 2x^2 + C
Solution: Separate: y dy = 2x dx. Integrate: y^2/2 = x^2 + C1, so y^2 = 2x^2 + C.
Question 6
Newton's Law of Cooling gives dT/dt = -k(T - 20) for an object cooling in a 20-degree room. The equilibrium temperature is:
A
20 degrees
B
0 degrees
C
-20 degrees
D
40 degrees
Solution: Equilibrium occurs where dT/dt = 0: T - 20 = 0, so T = 20 (the object approaches room temperature).
Question 7
The logistic differential equation differs from exponential growth because it includes:
A
a factor that limits growth as P approaches the carrying capacity
B
a negative growth rate
C
no proportionality constant
D
a second derivative
Solution: Logistic growth multiplies kP by (1 - P/M), which slows growth as P approaches the carrying capacity M.
Question 8
Solve dy/dx = 4x^3 with y(0) = -1.
A
y = x^4 - 1
B
y = x^4 + 1
C
y = 4x^4 - 1
D
y = x^4
Solution: Antidifferentiate: y = x^4 + C. Using y(0) = -1: 0 + C = -1, so y = x^4 - 1.
Question 9
If a quantity decays exponentially to 1/4 of its value in 6 hours, its half-life is:
A
3 hours
B
6 hours
C
1.5 hours
D
12 hours
Solution: Decaying to 1/4 means two successive halvings occurred in 6 hours, so each half-life is 6/2 = 3 hours.
Question 10
In the slope field for dy/dx = 2, every segment has:
A
the same slope of 2 (parallel segments)
B
slope 0
C
slopes that vary with x
D
slopes that vary with y
Solution: Since dy/dx = 2 is constant, every segment in the slope field is parallel with slope 2 (solutions are lines y = 2x + C).
Question 11
Solve dy/dx = 1/x (x > 0) with y(1) = 0.
A
y = ln(x)
B
y = ln(x) + 1
C
y = 1/x
D
y = -1/x^2
Solution: Antidifferentiate: y = ln(x) + C. Using y(1) = 0: ln(1) + C = 0, so C = 0 and y = ln(x).
Question 12
Solve dy/dx = cos(x) with y(0) = 1, then find y(pi/2).
A
2
B
1
C
0
D
pi/2
Solution: Antidifferentiate: y = sin(x) + C. y(0) = 1 gives C = 1, so y = sin(x) + 1. Then y(pi/2) = 1 + 1 = 2.
Question 13
A population follows P(t) = 500 e^(0.04t). What is the population growth rate dP/dt when t = 0?
A
20 per unit time
B
500 per unit time
C
0.04 per unit time
D
40 per unit time
Solution: dP/dt = 500(0.04)e^(0.04t) = 20 e^(0.04t). At t = 0, dP/dt = 20.
Question 14
Carbon-14 has a half-life of about 5730 years. The decay model uses k equal to:
A
(ln 2)/5730
B
5730 ln 2
C
ln(1/5730)
D
1/5730
Solution: From the half-life relation e^(-5730 k) = 1/2, we get k = (ln 2)/5730.
Question 15
For the logistic equation dP/dt = 0.5 P(1 - P/200), the population grows fastest when P equals:
A
100
B
200
C
50
D
0
Solution: Logistic growth is fastest at half the carrying capacity, P = M/2 = 200/2 = 100 (this is the inflection point of the solution curve).
Question 16
For dP/dt = kP(1 - P/M), the equilibrium solutions (where dP/dt = 0) are:
A
P = 0 and P = M
B
P = M/2 only
C
P = k
D
P = 1
Solution: dP/dt = 0 when P = 0 or when 1 - P/M = 0, i.e., P = M. These are the two equilibrium (constant) solutions.
Question 17
A tank holds 100 L of pure water. Brine with 2 g/L of salt flows in at 5 L/min (and well-mixed solution flows out at 5 L/min). If S is the salt amount, the model for dS/dt is:
A
dS/dt = 10 - (5S/100)
B
dS/dt = 2 - S
C
dS/dt = 10 - S
D
dS/dt = 5S
Solution: Salt in = (2 g/L)(5 L/min) = 10 g/min. Salt out = (concentration S/100)(5 L/min) = 5S/100. So dS/dt = 10 - 5S/100. (Mixing model adapted from the Vidaara JEE bank.)
Question 18
A savings account grows continuously at 4% per year: dA/dt = 0.04A. If A(0) = 1000, the balance after t years is:
A
1000 e^(0.04t)
B
1000(1.04)^t only
C
1000 + 0.04t
D
1000 e^(4t)
Solution: Continuous growth dA/dt = 0.04A has solution A = A0 e^(0.04t). With A0 = 1000, A(t) = 1000 e^(0.04t).
Question 19
Solve dy/dx = (y ln y)/x with the solution curve passing through (1, e).
A
y = e^x
B
y = x^e
C
y = e^(x^2)
D
y = x
Solution: Separate: dy/(y ln y) = dx/x. Let u = ln y; du = dy/y. Then du/u = dx/x, so ln|ln y| = ln|x| + C1, giving ln y = Cx. Through (1, e): ln e = C(1), so C = 1 and ln y = x, i.e., y = e^x. (Adapted from the Vidaara JEE bank.)
Question 20
For the slope field of dy/dx = -x/y, the solution curves are:
A
circles centered at the origin
B
straight lines
C
parabolas
D
exponential curves
Solution: Separating dy/dx = -x/y gives y dy = -x dx, so y^2/2 = -x^2/2 + C, i.e., x^2 + y^2 = C: circles centered at the origin.