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--- book:chap3:3.2_time_derivatives_of_vectors [2021/11/15 15:21] – [3.2 Time derivatives of vectors] abril
+++ book:chap3:3.2_time_derivatives_of_vectors [2021/11/18 02:03] (current) – [3.2.1 Self Test] jv
@@ Line 85: / Line 85: @@
     \frac{\mathrm{d}}{\mathrm{d} x} \ln  x &= x^{-1}
 \end{align*}
-Use only the three rules for derivatives
+and use only the three rules for derivatives
 \begin{align*}
     \frac{\mathrm{d}}{\mathrm{d} x} \bigl( f(x) + g(x) \bigr)
@@ Line 98: / Line 98: @@
 to work out the following derivatives
-**a)** $\displaystyle \sinh x = \frac{1}{2} \left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \quad$ and $\quad \displaystyle \cosh x = \frac{1}{2} \left( \mathrm{e}^x + \mathrm{e}^{-x} \right)$\\
+<WRAP group>
-**b)** $\cos x = \sin(\pi/2 + x) \qquad$\\
+<WRAP half column>
-**c)** $x^a = \mathrm{e}^{a \, \ln x}$ for $a \in \mathbb{R}$ What does this imply for the derivative of $f(x) = x^{-1}$?\\
+**a)** $\displaystyle \sinh x = \frac{1}{2} \left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \quad$ and $\quad \displaystyle \cosh x = \frac{1}{2} \left( \mathrm{e}^x + \mathrm{e}^{-x} \right)$
+**b)** $\cos x = \sin(\pi/2 + x) \qquad$
+**c)** $x^a = \mathrm{e}^{a \, \ln x}$ for $a \in \mathbb{R}$. \\
+$\quad$ What does this imply for the derivative of $f(x) = x^{-1}$?
+</WRAP>
+<WRAP half column>
 **d)** Use the result from c) to proof the quotient rule:
 \begin{align*}
     \frac{\mathrm{d}}{\mathrm{d} x} \frac{ f(x) }{ g(x) }
     = \frac{ f'(x) \, g(x) - f(x) \, g'(x) }{ \bigl( g(x) \bigr)^2 }
-\end{align*}\\
+\end{align*} \\
-**e)** $\displaystyle \tan x = \frac{ \sin x }{ \cos x } \quad $ and $\quad \displaystyle \tanh x = \frac{ \sinh x }{ \cosh x }$\\
+**e)** $\displaystyle \tan x = \frac{ \sin x }{ \cos x } \quad $ and $\quad \displaystyle \tanh x = \frac{ \sinh x }{ \cosh x }$ \\
-**e)** :!: Find the derivative of $\ln x$ solely based on
+**f)** :!: Find the derivative of $\ln x$ solely based on
 $\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} \mathrm{e}^x = \mathrm{e}^x$.
 \\
+$\quad$ ++Hint: | Use that $x = \mathrm{e}^{\ln x}$ and take the derivative of both sides.++
-++Hint: | Use that $x = \mathrm{e}^{\ln x}$ and take the derivative of both sides.++
+</WRAP>
+</WRAP>
@@ Line 120: / Line 129: @@
 In a moment we will also perform integrals to determine the work performed on a body
 when it is moving subject to a force.
-Practice you skills by evaluating the following integrals.
+Practice your skills by evaluating the following integrals.
+<WRAP group>
+<WRAP half column>
+**a)**   $\displaystyle \int_{-1}^1 {\mathrm{d} x}\, ( a + x )^2$ \\
+**b)**   $\displaystyle \int_{-5}^5 {\mathrm{d} q}\, ( a + b\, q^3 )$ \\
+**c)**   :!: $\displaystyle \int_0^B {\mathrm{d} k}\, \tanh^2( k x )$ \\
+**d)**   $\displaystyle \int_0^\infty {\mathrm{d} x}\, \mathrm{e}^{-x/L}$ \\
+**e)**   $\displaystyle \int_{-L}^L {\mathrm{d} y}\, \mathrm{e}^{-y/\xi}$
+</WRAP>
+<WRAP half column>
+**f)**   $\displaystyle \int_{0}^L {\mathrm{d} z}\, \frac{z}{a+b\,z^2}$ \\
+**g)**   $\displaystyle \int_{0}^\infty {\mathrm{d} x} \, x \, \mathrm{e}^{-x^2/(2Dt)}$ \\
+**h)**   $\displaystyle \int_{-\sqrt{Dt}}^{\sqrt{Dt}} {\mathrm{d} \ell} \, \ell \, \mathrm{e}^{-\ell^2/(2Dt)}$ \\
+**i)**   :!: $\displaystyle \int_{-\sqrt{Dt}}^{\sqrt{Dt}} {\mathrm{d} z}\, x \, \mathrm{e}^{-z\, x^2}$
+</WRAP>
+</WRAP>
-  -  $\displaystyle \int_{-1}^1 {\mathrm{d} x}\, ( a + x )^2$
-  -  $\displaystyle \int_{-5}^5 {\mathrm{d} q}\, ( a + b\, q^3 )$
-  -  :!: $\displaystyle \int_0^B {\mathrm{d} k}\, \tanh^2( k x )$
-  -  $\displaystyle \int_0^\infty {\mathrm{d} x}\, \mathrm{e}^{-x/L}$
-  -  $\displaystyle \int_{-L}^L {\mathrm{d} y}\, \mathrm{e}^{-y/\xi}$
-  -  $\displaystyle \int_{0}^L {\mathrm{d} z}\, \frac{z}{a+b\,z^2}$
-  -  $\displaystyle \int_{0}^\infty {\mathrm{d} x} \, x \, \mathrm{e}^{-x^2/(2Dt)}$
-  -  $\displaystyle \int_{-\sqrt{Dt}}^{\sqrt{Dt}} {\mathrm{d} \ell} \, \ell \, \displaystyle \mathrm{e}^{-\ell^2/(2Dt)}$
-  -  :!: $\displaystyle \int_{-\sqrt{Dt}}^{\sqrt{Dt}} {\mathrm{d} z}\, x \, \displaystyle \mathrm{e}^{-z\, x^2}$
 Except for the integration variable all quantities are considered to be constant.
 \\
-++Hint: | Sometimes symmetries can substantially reduce the work needed to evaluate an integral.++
+++Hint: | $\quad$
+Sometimes symmetries can substantially reduce the work needed to evaluate an integral.
+++
 ~~DISCUSSION|Questions, Remarks, and Suggestions~~