What is the equation of the line that is normal to $f(x)=cscx+cotx$ at $x=-pi/3$ ?

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Answer 1 · 2016-07-26T11:05:19

$3x-6y+pi-6sqrt3=0$ .

We know that $dy/dx=f'(x)$ gives the slope of tgt. to curve $C : y=f(x)$ at any pt. $P(x,y)$ on the curve.

since, normal line is $bot$ to tgt., its slope, at any pt. $P(x,y)$ , will be $-1/(f'(x)), if f'(x)!=0$

The Curve is $C : y=f(x)=cscx+cotx$

$:. f'(x)=-cscxcotx-csc^2x=-cscx(cotx+cscx)$

$:. f'(-pi/3)=-csc(-pi/3){cot(-pi/3)+csc(-pi/3)}$

$=-2/sqrt3(1/sqrt3+2/sqrt3)=-6/3=-2$

Therefore, the slope of normal $=1/2$ .

Also, $f(-pi/3)=csc(-pi/3)+cot(-pi/3)=-sqrt3$ .

Thus, the normal line passes thro. pt. $(-pi/3,-sqrt3)$ and has slope $=1/2$ .

Therefore, the eqn. of normal is given by,

$y+sqrt3=1/2(x+pi/3)$ , i.e., $6y+6sqrt3=3x+pi$ , or

$3x-6y+pi-6sqrt3=0$ .

What is the equation of the line that is normal to f(x)=cscx+cotx f(x)=cscx+cotx at x=-pi/3 x=-pi/3?