Question

Find the curve in the xy-plane that passes through the point each point is sec23x. nd whose slope at

Answer 1

Answer

Let the curve in the xy-plane be

Since the slope at each point of the curve is sec2(3x),

$\frac{\mathrm{d} y}{\mathrm{d} x}=\sec^2(3x)$

By integration, we will get

$y=\int\sec^2(3x)\, dx =\frac{\tan(3x)}{3}+C$

where, C is the constant of integration.

Hence, the equation of the curve is

$y =\frac{\tan(3x)}{3}+C$

Since the curve passes through the point ($\pi$/12, 2),

$2 =\frac{\tan\left ( 3\times \frac{\pi}{12} \right )}{3}+C$

$\Rightarrow 2 =\frac{\tan\left ( \frac{\pi}{4} \right )}{3}+C$

$\Rightarrow 2 =\frac{1}{3}+C$

$\Rightarrow C = 2-\frac{1}{3}=\frac{5}{3}$

Therefore, the equation is

${\color{Blue} y =\frac{\tan(3x)}{3}+\frac{5}{3}}$