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 \(F = \frac{d}{dt} \left( mv \right) .\)
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?
Solution
\( y = y_0 e ^{kt} \implies \frac{1}{2} = e^{140 k} \implies \frac{\ln{\frac{1}{2}}}{140} = k \implies 0.1 = e^{\frac{\ln {0.5}}{140}t} \implies t = 465.0699\)
Solve the differential equation \( \frac{dy}{dx} = 10 - 10y \) subject to the initial condition that \(x = 0\) when \(y = 2\).
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 \(y - 1 = e^{-10x} \implies y = e^{-10x} + 1\)
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 \(t = 5\).
Solution
\( \frac{dx}{dt} = kx \implies x = x_0e^{kt} \implies x = 2e^{kt} \implies 4 = 2e^{10k} \implies k = \frac{\ln{2}}{10} \implies x = 2e^{\frac{\ln{2}}{10}t} \implies x(5) = 2e^{\frac{\ln{2}}{2}} = 2\sqrt{2}.\)
Evaluate the following integrals:
\( \displaystyle \int x^2 \cos{x} \, dx \)
Hint
Tabular with \(x^2\) going down.Solution
\( x^2 \sin{x} + 2x \cos{x} - 2\sin{x} + C \)
\( \displaystyle \int x^2 \ln{x} \, dx \)
Hint
By parts with \(ln\) going down.Solution
\( \frac{x^3 \ln{x}}{3} - \frac{x^3}{9} + C \)
\( \displaystyle \int x^3 e^{4x} \, dx \)
Hint
Tabular with \(x^3\) going down.Solution
\( \frac{x^3 e^{4x}}{4} - \frac{3x^2 e^{4x}}{16} + \frac{6x e^{4x}}{64} - \frac{6 e^{4x}}{256} + C\)
\( \displaystyle \int 4x \sec^2{x} \, dx \)
Hint
By parts with \(\sec^2{x}\) going up.Solution
\( 4x \tan{x} - 4\ln{|\sec{x}|} + C \)
\( \displaystyle \int e^{-2x} \sin{(2x)} \, dx \)
Hint
Solve for unknown integral. Have \(\sin {(2x)}\) go down.Solution
\( -\frac{1}{4}e^{-2x}\sin{(2x)} -\frac{1}{4}e^{-2x}\cos{(2x)} + C \)
\( \displaystyle \int x \sqrt{1 - x} \, dx \)
Hint
Substitution. Let \(u = 1 - x\).Solution
\( -\frac{2}{3}(1 - x)^{\frac{3}{2}} + \frac{2}{5}(1 - x)^{\frac{5}{2}} + C \)