Study questions platform-wide or filter by specific tests with correct answers revealed.
According to Coulomb's Law, the force $F$ between two charges $q_1$ and $q_2$ separated by distance $r$ is:
$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$
Rearranging for permittivity:
$\epsilon_0 = \frac{q^2}{4\pi F r^2}$
Dimensional analysis:
- Charge $[q] = [AT]$
- Force $[F] = [MLT^{-2}]$
- Distance $[r] = [L]$
$[\epsilon_0] = \frac{[A^2 T^2]}{[M L T^{-2}] [L^2]} = [M^{-1} L^{-3} T^4 A^2]$
Let $I_n=\displaystyle\int\tan^n x\,dx$ for $n>1$. If $I_4+I_6 = a\tan^5x + bx^5 + C$, find the ordered pair $(a,b)$.
Use the reduction: $I_n+I_{n-2}=\displaystyle\int\tan^{n-2}x\cdot\tan^2x\,dx+\int\tan^{n-2}x\,dx=\int\tan^{n-2}x\sec^2x\,dx=\dfrac{\tan^{n-1}x}{n-1}+C$
$I_4+I_6$: set $n=5$: $I_4+I_6=\dfrac{\tan^5x}{5}+C$
Comparing with $a\tan^5x+bx^5+C$: $a=\dfrac{1}{5}$, $b=0$.
$(a,b) = \left(\dfrac{1}{5},0\right)$
Let $p = \displaystyle\lim_{x\to0^+}\left(1+\tan^2\sqrt{x}\right)^{\frac{1}{2x}}$. Find $\ln p$.
This is a $1^\infty$ indeterminate form. Take logarithm:
$\ln p = \lim_{x\to0^+}\dfrac{\ln(1+\tan^2\sqrt{x})}{2x}$
Let $t=\sqrt{x}$, so $x=t^2$, $x\to0^+$ means $t\to0^+$:
$= \lim_{t\to0^+}\dfrac{\ln(1+\tan^2 t)}{2t^2} = \lim_{t\to0^+}\dfrac{\tan^2 t}{2t^2} = \dfrac{1}{2}$
(using $\ln(1+u)\approx u$ for small $u$ and $\lim_{t\to0}\frac{\tan t}{t}=1$)
Sign in to join the conversation and share your thoughts.
Log In to Comment