Compute a vector valued integral

by RoRi

June 1, 2016

Compute the vector valued integral

$\int_0^1 (te^t \mathbf{i} + t^2 e^t \mathbf{j} + te^{-t} \mathbf{k}) \, dt.$

We compute

$\begin{align*} \int_0^1 (te^t \mathbf{i} + t^2 e^t \mathbf{j} + te^{-t} \mathbf{k} ) \, dt &= \left( \int_0^1 te^t \, dt, \int_0^1 t^2 e^t \, dt, \int_0^1 te^{-t} \, dt \right) \\[9pt] &= \left( 1, e-2, 1 - \frac{2}{e} \right). \end{align*}$

Compute the vector valued integral

by RoRi

June 1, 2016

Compute the vector valued integral

$\int_0^1 \left( \frac{e^t}{1+e^t} \mathbf{i} + \frac{1}{1+e^t} \mathbf{j} \right) \, dt.$

We compute,

$\begin{align*} \int_0^1 \left( \frac{e^t}{1+e^t} \mathbf{i} + \frac{1}{1+e^t} \mathbf{j} \right) \,dt &= \left( \int_0^1 \frac{e^t}{1+e^t} \, dt \right) \mathbf{i} + \left( \int_0^1 \frac{1}{1+e^t} \, dt \right) \mathbf{j} \\[9pt] &= \log \left( \frac{1+e}{2} \right) \mathbf{i} + \left( 1 - \log \frac{1+e}{2} \right) \mathbf{j}. \end{align*}$

Compute the integral of (sin t, cos t, tan t)

by RoRi

May 31, 2016

Compute the vector-valued integral

$\int_0^{\frac{\pi}{4}} (\sin t, \cos t, \tan t) \, dt.$

We compute

$\begin{align*} \int_0^{\frac{\pi}{4}} (\sin t, \cos t, \tan t) \, dt &= \left( \int_0^{\frac{\pi}{4}} \sin t \, dt, \int_0^{\frac{\pi}{4}} \cos t \, dt, \int_0^{\frac{\pi}{4}} \tan t \, dt \right) \\[9pt] &= \left( 1 - \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, - \log \frac{\sqrt{2}}{2} \right). \end{align*}$

Compute the integral of (t, t^1/2, e^t)

by RoRi

May 31, 2016

Compute the vector-valued integral

$\int_0^1 (t, \sqrt{t}, e^t) \, dt.$

We compute,

$\begin{align*} \int_0^1 (t, \sqrt{t}, e^t) \, dt &= \left( \int_0^1 t \, dt , \int_0^1 \sqrt{t} \, dt, \int_0^1 e^t \, dt \right) \\[9pt] &= \left( \frac{1}{2}, \frac{2}{3}, e-1 \right). \end{align*}$

Prove that the angle between F(t) and F'(t) is constant for a given vector function

by RoRi

May 31, 2016

Consider the vector-valued function

$F(t) = \frac{2t}{1+t^2} \mathbf{i} + \frac{1-t^2}{1+t^2} \mathbf{j} + \mathbf{k}.$

Prove that the angle between $F(t)$ and $F'(t)$ is a constant.

Proof. First, we have

$F(t) = \left( \frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}, 1 \right) \quad \implies \quad F'(t) = \left( \frac{2-2t^2}{(1+t^2)^2}, \frac{-4t}{(1+t^2)^2}, 0 \right).$

Thus,

$\begin{align*} F(t) \cdot F'(t) &= \left( \frac{2t}{1+t^2} \right) \left( \frac{2-2t^2}{(1+t^2)^2} \right) + \left( \frac{1-t^2}{1+t^2} \right) \left( \frac{-4t}{(1+t^2)^2} \right) \\[9pt] &= \frac{4t-4t^3 - 4t + 4t^3}{(1+t^2)^3} \\[9pt] &= 0. \end{align*}$

Hence, $F(t)$ and $F'(t)$ are orthogonal, so the angle between them is constant, $\theta = \frac{\pi}{2}. \qquad \blacksquare$

Compute derivatives of F(t) = log (1+t²) i + arctan t j + 1/(1+t²) k

by RoRi

May 31, 2016

Compute the derivatives $F'(t)$ and $F''(t)$ of the vector valued function

$F(t) = \log(1+t^2) \mathbf{i} + \arctan t \mathbf{j} + \frac{1}{1+t^2} \mathbf{k}.$

We compute

$\begin{align*} F'(t) &= \left( \frac{2t}{1+t^2}, \frac{1}{1+t^2}, \frac{-2t}{(1+t^2)^2} \right) \\[9pt] F''(t) &= \left( \frac{2(1-t^2)}{(1+t^2)^2}, \frac{-2t}{(1+t^2)^2}, \frac{6t^2 - 2}{(1+t^2)^2} \right). \end{align*}$

Compute derivatives of F(t) = cosh t i + sinh (2t) j + e^-3tk

by RoRi

May 31, 2016

Compute the derivatives $F'(t)$ and $F''(t)$ of the vector valued function

$F(t) = \cosh t \mathbf{i} + \sinh (2t) \mathbf{j} + e^{-3t} \mathbf{k}.$

We compute

$\begin{align*} F'(t) &= \sinh t \mathbf{i} + 2 \cosh (2t) \mathbf{j} - 3e^{-3t} \mathbf{k} \\ F''(t) &= \cosh t \mathbf{i} + 4 \sinh (2t) \mathbf{j} + 9e^{-3t} \mathbf{k}. \end{align*}$

Compute derivatives of F(t) = 2e^ti + 3e^tj

by RoRi

May 31, 2016

Compute the derivatives $F'(t)$ and $F''(t)$ of the vector valued function

$F(t) = 2e^t \mathbf{i} + 3e^t \mathbf{j}.$

We compute,

$\begin{align*} F'(t) &= 2e^t \mathbf{i} + 3e^t \mathbf{j} \\ F''(t) &= 2e^t \mathbf{i} + 3e^t \mathbf{j}. \end{align*}$

Compute derivatives of F(t) = (arcsin t, arccos t)

by RoRi

May 31, 2016

Compute the derivatives $F'(t)$ and $F''(t)$ where $F(t)$ is the vector valued function

$F(t) = (\arcsin t, \arccos t).$

We compute

$\begin{align*} F'(t) &= \left( \frac{1}{\sqrt{1-t^2}}, \frac{-1}{\sqrt{1-t^2}} \right) \\ F''(t) &= \left( \frac{t}{(1+t^2)^{\frac{3}{2}}}, \frac{-t}{(1+t^2)^{\frac{3}{2}}} \right). \end{align*}$