Question

Normal and tangent lines at a point.how to solve the tangent,normal line and slope of normal line?someone pls explain the details to me.

Answer 1

tangent: take derivative, plug in x - value, that gives slope. use point slope formula

Answer 2

normal, same thing except this time the slope is the negative reciprocal of the tangent slope

Answer 3

detailed enough? an example might help

Answer 4

can you solve this pls make it a detailed.
y=x^3-2x=1 at (1,2)

Answer 5

satellite can you solve this pls make it a detailed.
y=x^3-2x=1 at (1,2)

Answer 6

Do you know how to take the derivative dy/dx of your equation?

Answer 7

yup its 3x^2-2 right?

Answer 8

y=x^3-2x+1 at (1,2)

Answer 9

Good. That is the slope of your curve.

Answer 10

yesh. now replace x by 1 to get your slope, use point slope formula to find the line

Answer 11

@phi not to be argumentative but
\[y'=3x^2-2\] is not a slope

Answer 12

At each x point, it's different because the curve is curving... it's not a straight line.
So at x =1 you can find the slope using the derivative.

Answer 13

The Formula of the Tangent is
\[t(x) = f(a)+f'(a)(x-a)\]
The normal is ortogonal to the constant, then it may have the opposite slope. Then it is
\[t(x) = f(a)-f'(a)(x-a)\]

Answer 14

it is the formula for the slope. the slope is a number. you get it by replacing x by whatever number you have for x

Answer 15

@aoigetsu that is just not right, sorry.

Answer 16

It isn't? Why?

Answer 17

normal line is perpendicular.

Answer 18

at any rate for this problem the derivative is
\[3x^2-2\] put x = 1 get
\[3-2=1\] so slope is 1

Answer 19

you have the point (1,2) the slope is 1, so you can find the equation of the tangent line via
\[y-2=x-1\]
\[y=x+1\]

Answer 20

Oh yeah
Sorry, bad english

Try it out:
http://graphsketch.com/
The function is x², the derivate is 2x and the point I choose is x=1
Then the functions that will express this relation using the formulas is

f(x) = x²
tangent=2x-1
normal = -2x+1

Answer 21

can you pls show how did you get the tangent and normal line?

Answer 22

now for normal line in this case it is easy because your slope is just 1, so perpendicular line has slope -1

Answer 23

ok

Answer 24

you have the point. it is (1,2)

Answer 25

you find the slope by replacing x in the derivative by 1

Answer 26

you get
\[3\times 1^2-2=3-2=1\]

Answer 27

so your slope is 1. then you use the point slope formula
\[y-y_1=m(x-x_1)\] with
\[m=1, x_1=1, y_1=2\]

Answer 28

you get
\[y-2=1(x-1)\]
\[y-2=x-1\]
\[y=x+1\]

Answer 29

this is a lousy example because both the x coordinate and the slope are 1, so you miss some work you would otherwise have to do

Answer 30

but that is your answer with probably more steps than necessary but i wanted to write them all out in any case

Answer 31

clear or no?

Answer 32

am how do you get the tangent?

Answer 33

you mean the tangent line or the slope of the tangent line?

Answer 34

the tangent line

Answer 35

and the normal line?

Answer 36

ok lets start again with a different problem. maybe will make it clearer. different but easy

Answer 37

ok pls

Answer 38

\[f(x)=x^2-4\] at the point (3,5)

Answer 39

notice that (3,5) is on the graph because
\[f(3)=3^2-4=5\] so really i could have just said the 3, and you could have come up with the 5

Answer 40

now what is
\[f'(x)\]?

Answer 41

am you did not derive it?

Answer 42

\[f(x)=x^2-4\] what is
\[f'(x)\]?

Answer 43

we need that one. it is pretty easy in this case yes?

Answer 44

2x

Answer 45

got it.

Answer 46

now i want the slope of the line tangent to the graph of
\[y=x^2-4\] at the point (3,5)
how do i find it?

Answer 47

i find it by taking
\[f'(3)\]

Answer 48

\[f'(x)=2\times 3=6\] and that is the slope i am looking for

Answer 49

so now i want an equation for the line tangent to the graph. i have a point, namely (3,5)
and i have a slope, it is 6

Answer 50

so it is straight forward to find the equation for the line. i use the point slope formula and write
\[y-5=6(x-3)\]

Answer 51

some algebra gives me
\[y-5=6x-18\]
\[y=6x-13\] and that is it, that is the line tangent to the graph at (3,5)

Answer 52

that is a worked out procedure for finding the tangent line. now on to the "normal" line. we have done most of the work

Answer 53

the tangent line has slope 6. the normal line is perpendicular, so it has slope
\[-\frac{1}{6}\]

Answer 54

and again we use the point slope formula to find the equation

Answer 55

am after the tangent line our instructor get T is it the tangent how do you get it?he then solve for the normal line

Answer 56

i guess its the equation

Answer 57

the tangent equation

Answer 58

ok we have the slope of the tangent, it is 6, so we find the slope of the normal by taking the negative reciprocal. it is
\[-\frac{1}{6}\]

Answer 59

the use the point slope formula again
\[y-5=-\frac{1}{6}(x-3)\] etc

Answer 60

you do not use the tangent line to find the normal line. you use the point - slope formula with the negative reciprocal of the slope of the tangent line

Answer 61

thank u very much for helping me.

Answer 62

yw

Answer 63

you really helped me a lot

Answer 64

glad to