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Find the equation of the tangent and normal to the curves.$ \;y=\large\frac{1+\sin x}{\cos x}$ at $\;x=\large\frac{\pi}{4}$

Note: This is part 4th of a 4 part question, split as 4 separate questions here.

1 Answer

  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
$y=\large\frac{1+\sin x}{\cos x}$ at $x=\large\frac{\pi}{4}$
Step 1:
When $x=\large\frac{\pi}{4}$$ y=\large\frac{1+\Large\frac{1}{\sqrt 2}}{\Large\frac{1}{\sqrt 2}}$
$\qquad=\sqrt 2+1$
We are required to find the equations of the tangent and the normal at $\bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$
Step 2:
The slope of the tangent at $\bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$ is
$m=\large\frac{dy}{dx_{(\Large\frac{\pi}{4},\sqrt 2+1)}}$
$\large\frac{dy}{dx}=\large\frac{\cos x(\cos x)-(1+\sin x)(-\sin x)}{\cos ^2 x}$
$\qquad=\large\frac{\cos^2 x+\sin x+\sin^2 x}{\cos ^2 x}$
$\qquad=\large\frac{1+\sin x}{\cos ^2 x}$
$m=\large\frac{dy}{dx_{(\Large\frac{\pi}{4},\sqrt 2+1)}}$
$\quad=\large\frac{1+\Large \frac{1}{\sqrt 2}}{\Large\frac{1}{2}}$
$\quad=\large\frac{\sqrt 2+1}{\sqrt 2}$$.2$
$\quad=2 +\sqrt 2$
Equation of the tangent at $\bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$
$y-(\sqrt 2+1)=(2 +\sqrt {2})(x-\large\frac{\pi}{4})$
Step 3:
Normal is $\perp$ to the tangent. Its slope equation of the normal at $\bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$
$y-(\sqrt 2+1)=\large\frac{-1}{(2 +\sqrt 2)}($$x-\large\frac{\pi}{4})$
answered Jul 23, 2013 by meena.p
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