Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App