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

- anonymous

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

- katieb

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- dumbcow

derivative of cos is -sin
--> dy/dx = -10sin x

- anonymous

how will i find these two also
The slope of the tangent line is m1= . The equation of the tangent line is y= .

- dumbcow

slope = dy/dx evaluated at some x

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- anonymous

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

- dumbcow

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

- anonymous

cos/-sin is the solution?

- dumbcow

??
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

- anonymous

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?

- dumbcow

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

- anonymous

-10sin(pi/3) ok

- dumbcow

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

- anonymous

60degrees (1/2,sqrt3/2

- anonymous

what do i do with those numbers

- dumbcow

which one refers to sin

- anonymous

sqrt3/2

- dumbcow

cos() --> similar to x coord
sin() --> similar to y coord
good
so slope is -5sqrt(3)

- anonymous

ok

- anonymous

the line tangent?

- dumbcow

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

- anonymous

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

- dumbcow

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

- anonymous

haha yeah

- anonymous

my homework software accepted that

- dumbcow

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)

- anonymous

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

- anonymous

wait

- dumbcow

m2?? haha, same function different point

- anonymous

my software didnt accept the previous answer

- dumbcow

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

- anonymous

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

- dumbcow

well what does that number represent ?

- dumbcow

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

- anonymous

hahaha

- anonymous

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

- anonymous

when i simplify

- dumbcow

good

- dumbcow

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

- anonymous

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

- dumbcow

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

- anonymous

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

- dumbcow

no

- dumbcow

y = mx+b
b = y - mx
m = -10sin(pi/3)
(x,y) = (pi/3,5)
b = 5 -(-10sin(pi/3)*pi/3)

- anonymous

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

- dumbcow

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

- anonymous

ohhhhh ok

- anonymous

im not sure how you solve for b here

- dumbcow

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

- anonymous

omg

- anonymous

i didnt see the y-mx

- anonymous

i got 45

- dumbcow

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

- dumbcow

i get 14.069

- anonymous

my calc was in radians mode

- dumbcow

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

- anonymous

ahh i didnt distribute the negative outside the parantheisis

- anonymous

whats next

- dumbcow

enter the equation of tangent line
y = ?

- dumbcow

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

- anonymous

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

- dumbcow

i dunno what they mean by m2

- dumbcow

is their another function, or another point ??
does it look like dy^2/dx

- anonymous

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

- dumbcow

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

- dumbcow

this will be easier
for perpendicular lines: m1*m2 = -1
m2 = -1/m1

- dumbcow

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

- dumbcow

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

- anonymous

what is that link for?

- anonymous

thanks for the help

- dumbcow

your welcome
link is to check your answer

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