BC Calculus: Extra Review Problems for Chapter 6



  1. Suppose that a body of mass \(m\) is moving in a straight line with velocity \(v\) encounters a resistance proportional to the velocity and that this is the only force acting on the body. If the body starts with velocity \(v_0\), how far does it travel in time \(t\)? Assume \(F = \frac{d}{dt} \left( mv \right) .\)
  2. Solution \( \frac{d}{dt} \left( mv \right) = kv \implies \displaystyle \int \frac{dv}{v} = \int \frac{k}{m} \, dt \implies \ln{|v|} = \frac{k}{m}t + C \implies |v| = Ce^{\frac{kt}{m}} \), and \( v = v_0e^{\frac{kt}{m}} \implies \int_0^t v_0 e^{\frac{kx}{m}} \, dx = \left. \frac{v_0 m}{k} e^{\frac{kx}{m}} \right|_{0}^{t} = \frac{v_0m}{k} \left( e^{\frac{kt}{m}} - 1 \right) \)
  3. Question two has been intentionally left blank.
  4. Solution Really? Get a life.
  5. The half-life of polonium is 140 days, but your sample will not be useful to you after 90% of the radioactive nuclei originally present have disintegrated. About how many days can you use the polonium?
  6. Solution \( y = y_0 e ^{kt} \implies \frac{1}{2} = e^{140 k} \implies \frac{\ln{\frac{1}{2}}}{140} = k \implies 0.1 = e^{\frac{\ln {0.5}}{140}t} \implies t = 465.0699\)
  7. Solve the differential equation \( \frac{dy}{dx} = 10 - 10y \) subject to the initial condition that \(x = 0\) when \(y = 2\).
  8. Solution \( \displaystyle \int \frac{dy}{1 - y} = \int 10 \, dx \implies \ - \ln {|1 - y|} = 10x + C \implies \ln{|1 - y|} = -10x + C \implies |1 - y| = Ce^{-10x}.\) The initial condition is (0, 2), so \( C = 1\) and \(y - 1 = e^{-10x} \implies y = e^{-10x} + 1\)
  9. The velocity of a particle moving along the x-axis is proportional to \(x\). This is to say the particle's velocity is proportional to its displacement. At time \(t = 0\), the particle is located at \(x = 2\), and at time \(t = 10\), it is at \(x = 4\). Find its position at \(t = 5\).
  10. Solution \( \frac{dx}{dt} = kx \implies x = x_0e^{kt} \implies x = 2e^{kt} \implies 4 = 2e^{10k} \implies k = \frac{\ln{2}}{10} \implies x = 2e^{\frac{\ln{2}}{10}t} \implies x(5) = 2e^{\frac{\ln{2}}{2}} = 2\sqrt{2}.\)
  11. Evaluate the following integrals:
    1. \( \displaystyle \int x^2 \cos{x} \, dx \)
    2. Hint Tabular with \(x^2\) going down.
      Solution \( x^2 \sin{x} + 2x \cos{x} - 2\sin{x} + C \)
    3. \( \displaystyle \int x^2 \ln{x} \, dx \)
    4. Hint By parts with \(ln\) going down.
      Solution \( \frac{x^3 \ln{x}}{3} - \frac{x^3}{9} + C \)
    5. \( \displaystyle \int x^3 e^{4x} \, dx \)
    6. Hint Tabular with \(x^3\) going down.
      Solution \( \frac{x^3 e^{4x}}{4} - \frac{3x^2 e^{4x}}{16} + \frac{6x e^{4x}}{64} - \frac{6 e^{4x}}{256} + C\)
    7. \( \displaystyle \int 4x \sec^2{x} \, dx \)
    8. Hint By parts with \(\sec^2{x}\) going up.
      Solution \( 4x \tan{x} - 4\ln{|\sec{x}|} + C \)
    9. \( \displaystyle \int e^{-2x} \sin{(2x)} \, dx \)
    10. Hint Solve for unknown integral. Have \(\sin {(2x)}\) go down.
      Solution \( -\frac{1}{4}e^{-2x}\sin{(2x)} -\frac{1}{4}e^{-2x}\cos{(2x)} + C \)
    11. \( \displaystyle \int x \sqrt{1 - x} \, dx \)
    12. Hint Substitution. Let \(u = 1 - x\).
      Solution \( -\frac{2}{3}(1 - x)^{\frac{3}{2}} + \frac{2}{5}(1 - x)^{\frac{5}{2}} + C \)