Question

Find the cosine of angle of intersection of curves $y=2^{x}lnx$ and $y=x^{2x}-1$ at $(1,0)$.

Answer 1

Curves are $y=2^x\ln x$ and $y=x^{2x}-1$ at $(1,0)$

${\left(\frac{dy}{dx}\right)}_{curve1}$:

$\frac{dy}{dx}=\frac{2^x}{x}+2^x\ln 2\ln x$

${\left(\frac{dy}{dx}\right)}_{at(1,0)}=\frac{2}{1}+0=2$

${\left(\frac{dy}{dx}\right)}_{curve2}$:

Let, $y_1=x^{2x}$

Taking $\ln$ both sides,

$\ln y_1=2x\ln x$

Differentiating with respect to $x$,

$\frac{1}{y_1}\frac{d{y}_1}{dx}=2(x\frac{1}{x}+1\ln x)$

$\frac{d{y}_1}{dx}=2(1+\ln x)x^{2x}\{y_1=x^{2x}\}$

${\left(\frac{dy}{dx}\right)}_{curv{e}_2}$:

$\frac{dy}{dx}=2(1+\ln x)x^{2x}$

${\left(\frac{dy}{dx}\right)}_{at(1,0)}=2(1+\ln 0)1^1=2$

$\therefore$ Curves parallel to each other at $(1,0)$

$\tan \theta =0^o$

$\cos \theta =$ any real number.