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For $x\in\left(0,\dfrac{1}{4}\right)$, the derivative of $\tan^{-1}\!\left(\dfrac{6x\sqrt{x}}{1-9x^3}\right)$ equals $\sqrt{x}\cdot g(x)$. Find $g(x)$.
Write $\frac{6x\sqrt{x}}{1-9x^3} = \frac{2\cdot3x^{3/2}}{1-(3x^{3/2})^2}=\frac{2t}{1-t^2}$ where $t=3x^{3/2}$.
So $\tan^{-1}\!\left(\frac{2t}{1-t^2}\right)=2\tan^{-1}(t)=2\tan^{-1}(3x^{3/2})$
Differentiate: $\frac{d}{dx}[2\tan^{-1}(3x^{3/2})] = \frac{2}{1+9x^3}\cdot 3\cdot\frac{3}{2}x^{1/2} = \frac{9\sqrt{x}}{1+9x^3}$
Since this equals $\sqrt{x}\cdot g(x)$: $g(x)=\dfrac{9}{1+9x^3}$
A projectile is thrown at angles $(45°-\theta)$ and $(45°+\theta)$ with the horizontal. The ratio of their horizontal ranges is:
Using $R = \dfrac{u^2\sin2\alpha}{g}$:
$R_1 = \dfrac{u^2\sin(90°-2\theta)}{g} = \dfrac{u^2\cos2\theta}{g}$
$R_2 = \dfrac{u^2\sin(90°+2\theta)}{g} = \dfrac{u^2\cos2\theta}{g}$
$R_1 : R_2 = \mathbf{1:1}$
Energy flux (Intensity) is the rate of heat flow per unit area:
$J = \frac{H}{A} = \frac{k \Delta T}{L}$
$J = \frac{0.1 \times (10^3 - 10^2)}{1} = 0.1 \times 900 = 90 \text{ Wm}^{-2}$
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