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At what angle $\theta $ do the curves $y=a^{x}$ and $y=b^{x}$ intersect $(a{\neq}b)$?

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Toolbox:
  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
  • If the curves $y=f(x),y=g(x)$ intersect at $(x_1,y_1)$ then the angle of intersection of the curves is the angle between the tangents at the point $(x_1,y_1)$
  • If $m_1=t'(x_1)$ and $m_2=g'(x_1)$ then the angle is $\tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2}$ when the curves are orthogonal , the tangents are perpendicular to each other and $m_1m_2=-1$
Step 1:
Let $y=a^x$----(i) and $y=b^n $----(ii) intersect at $(x_1,y_1)$
Step 2:
$\large\frac{dy}{dx}$$=a^x \log a$ and $\large\frac{dy}{dx}$$=b ^x \log b$ on differentiating (i) and (ii)
Step 3:
The slopes of the tangents to the curves at $(x_1,y_1)$ are $m_1=a^{x_1} \log b$ and $m_2=b^{x_1} \log b$
The angle between the curves is
$\tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2}=\frac{a^{x_1}\log a -b^{x_1}\log b}{1+ a ^{x_1} \log a b^{x_1}\log b}$
Step 4:
Now $y_1=a^{x_1}$ and $y_1=b^{x_1}$
Therefore the angle is $\tan ^{-1} \large\frac{y_1 \log a-y_1 \log b}{1+{y_1}^2 \log a \log b}$
$a^{x_1}=b^{x_1}$ but $a \neq b=>x_1=0$
Therefore $y_1=1$
The angle is $\tan ^{-1} \large\frac{\log a-\log b}{1+\log a \log b}$
answered Jul 24, 2013 by meena.p
 

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