1.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 4
V =
2.IntegralsHard
The region between the line y = 3x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 3xy = x²
V =
3.Sequences & seriesMedium
The power series Σk=1∞ (2k)!·(x − 3)ᵏ/(k!)² converges on an interval around x₀ = 3. Find the right end x of that interval.
Σ (2k)!·(x − 3)ᵏ/(k!)²x₀ = 3
x =
4.Sequences & seriesHard
f(x) = sin x is approximated by its Maclaurin polynomial of degree 3 at x = 0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₃|.
f(x) = sin xn = 3x = 0.5M = 1
|R₃| =
5.Multivariable calculusMedium
For f(x, y) = 4x²y − 4x³ + 3xy, find the maximum rate of change of f at the point (1, 2), that is |∇f|.
f(x, y) = 4x²y − 4x³ + 3xyP = (1, 2)
|∇f| =
6.Multivariable calculusHard
Find the critical points of f(x, y) = −x³ + 3x − y² − 4y, classify them, and give the value fmax of f at its local maximum.
f(x, y) = −x³ + 3x − y² − 4y
fmax =
7.Differential equationsMedium
Solve y″ + 6y′ + 5y = 0 with y(0) = 5 and y′(0) = 2. Find y(2).
y″ + 6y′ + 5y = 0y(0) = 5y′(0) = 2
y(2) =
8.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 32.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 32.5y(0) = 0y′(0) = 0
ymax =