Question

Find the equation of the line normal to the curve y=secx+x at the point (0,1)

Answer 1

i dont understand what should i do first ??

Answer 2

the normal slope is opposite reciprocal of the tangent slope. So just find the tangent slope and then use point slope form

Answer 3

what should i do ?

Answer 4

do you know how to take derivative?

Answer 5

\[\frac{ dy }{ dx }= \sec x \tan x + 1\]

put x=0, we get slope of tangent equal to '1'.

so the slope of normal is equal to (-1/1)=-1

hence equation of normal is \[(y-1)=-1(x) \]

so the equation of normal is \[y=1-x\]

Answer 6

if I read that correctly, you're just being asked to find the equation of the line tangent to sec(x)+x

which means the equation of the tangent line to it, which means as sourwing said, find the slope, then use the given point of (0,1) to get the equation of the line

Answer 7

ok so the slope is and i dont understand what u did after that

Answer 8

find the derivative. That's what you need to do. Do you know how?

Answer 9

yep is it secxtanx+1?

Answer 10

whats next?

Answer 11

plug 0 in for x

Answer 12

\(\bf \cfrac{d}{dx}[sec(x)+x]\implies f'(x)=sec(x)tan(x)+1\qquad ({\color{blue}{ 0}},1)
\\ \quad \\
f'(x)=sec({\color{blue}{ 0}})tan({\color{blue}{ 0}})+1\)

Answer 13

that'd give you the value of the slope
recall that a derivative is nothing but the EQUATION to find the slope

Answer 14

which equals 1. and so the normal slope is -1/1 = -1

answer:
y - 0 = -1(x-1)

Answer 15

\(\bf f'(x)=sec({\color{blue}{ 0}})tan({\color{blue}{ 0}})+1\implies f'(x)=\cfrac{1}{cos(0)}\cdot \cfrac{sin(0)}{cos(0)}+1\)

Answer 16

notice your Unit Circle, the values for cosine and sine at 0

Answer 17

what did u use to show -1/1??

Answer 18

it's the opposite reciprocal of 1

Answer 19

hhyudg he's using the PERPENDICULAR LINE to THAT TANGENT line, also called as the NORMAL LINE

the slope of a line perpendiicular to another is, NEGATIVE RECIPROCAL to that one
meaning say if you had a slope of \(\bf \cfrac{1}{{\color{blue}{ 1}}}\qquad reciprocal\implies \cfrac{{\color{blue}{ 1}}}{1}\qquad negative\implies -\cfrac{{\color{blue}{ 1}}}{1}\implies -1\)