Question

y=10cosx at the point (π/3,5), what is the derivative

Answer 1

derivative of cos is -sin

--> dy/dx = -10sin x

Answer 2

how will i find these two also

The slope of the tangent line is m1= . The equation of the tangent line is y= .

Answer 3

slope = dy/dx evaluated at some x

Answer 4

im not familiar with that method. would you mind demonstrating it please?

Answer 5

what method?
its just the rule, derivative of sin is cos
derivative of cos is -sin

Answer 6

cos/-sin is the solution?

Answer 7

??

your answer will have 3 parts
a) derivative, dy/dx
b) slope of tangent line, take dy/dx plug in x=pi/3
c) equation of tangent line, use slope and find y_intercept

Answer 8

i dont understand how to do any of those. im working ahead on my homework b/c im going to busy with other things. so my teacher hasnt cover those yet and i am clueless. can you help me out please?

Answer 9

i already gave you the derivative,
now just plug in pi/3 for x to get slope

Answer 10

-10sin(pi/3) ok

Answer 11

yep...now use unit circle or calculator to find out what sin(pi/3) is

Answer 12

60degrees (1/2,sqrt3/2

Answer 13

what do i do with those numbers

Answer 14

which one refers to sin

Answer 15

sqrt3/2

Answer 16

cos() --> similar to x coord
sin() --> similar to y coord

good
so slope is -5sqrt(3)

Answer 17

ok

Answer 18

the line tangent?

Answer 19

y = mx+b

m = -5sqrt3
(x,y) = (pi/3,5)
find b

Answer 20

wait, i didnt mention that my homework software was satisfied with -10sinsqrt(3) as the slope

Answer 21

you mean...-10sin(pi/3)

Answer 22

haha yeah

Answer 23

my homework software accepted that

Answer 24

ok, same thing...you can interchange sqrt3/2 and sin(pi/3) because they are equal

y = mx+b

b = y - mx

b = 5 -(-10sin(pi/3)*pi/3)

Answer 25

ok thanks now i have to do slope and tangent line for m2

Answer 26

wait

Answer 27

m2?? haha, same function different point

Answer 28

my software didnt accept the previous answer

Answer 29

what did you put in?
did you enter it as y=mx+b form

Answer 30

i entered it in this form 5 -(-10sin(pi/3)*pi/3)

Answer 31

well what does that number represent ?

Answer 32

i was trying to help you, i didn't give you the final answer

Answer 33

hahaha

Answer 34

5 -(-10sin(pi/3)*pi/3)

i got 5+10sin(pi/3)pi/3

Answer 35

when i simplify

Answer 36

good

Answer 37

not done yet...what did you just solve for?
what answer are they looking for..."equation of tangent line"

Answer 38

the line that runs diagonally and only touches a parabola on its sides

Answer 39

yes thats definition of tangent line.

equation of tangent line is : y = mx+b
so answer must be in that form: slope*x + y_intercept

Answer 40

our slope is 10sin(pi/3) and y-int is 5 right?

Answer 41

no

Answer 42

y = mx+b

b = y - mx

m = -10sin(pi/3)
(x,y) = (pi/3,5)

b = 5 -(-10sin(pi/3)*pi/3)

Answer 43

i dont see how it isnt. you have m=-10sin(pi/3) and y=5

Answer 44

the y_coord of the point is 5.
don't confuse that with y_intercept, where line crosses y-axis
we represent y_intercept as b

Answer 45

ohhhhh ok

Answer 46

im not sure how you solve for b here

Answer 47

b = y - mx

b = 5 -(-10sin(pi/3)*pi/3)

Answer 48

omg

Answer 49

i didnt see the y-mx

Answer 50

i got 45

Answer 51

really? after multiplying by pi....or did you round it to 45

Answer 52

i get 14.069

Answer 53

my calc was in radians mode

Answer 54

yeah thats what you want
you could probably leave it as:
b= 5+10(pi/3)*sin(pi/3),

Answer 55

ahh i didnt distribute the negative outside the parantheisis

Answer 56

whats next

Answer 57

enter the equation of tangent line

y = ?

Answer 58

http://www.wolframalpha.com/input/?i=y+%3D+10cos%28x%29+find+tangent+line+at+x%3Dpi%2F3

here is a nice reference to check your answers

Answer 59

ok thanks m2, do find the derivative of the derivative of m1

Answer 60

i dunno what they mean by m2

Answer 61

is their another function, or another point ??

does it look like dy^2/dx

Answer 62

The slope of the normal line is m2
The equation of the normal line is y

Answer 63

oh ok, so m1 refers to tangent line
m2 refers to normal line which is perpendicular to tangent line at point (pi/3, 5)

Answer 64

this will be easier

for perpendicular lines: m1*m2 = -1
m2 = -1/m1

Answer 65

Then find equation of normal line same way
solve for y_intercept
b = y - m2*x

Answer 66

http://www.wolframalpha.com/input/?i=y+%3D+10cos%28x%29+find+normal+line+at+x%3Dpi%2F3

Answer 67

what is that link for?

Answer 68

thanks for the help

Answer 69

your welcome

link is to check your answer