anonymous
  • anonymous
y=10cosx at the point (π/3,5), what is the derivative
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dumbcow
  • dumbcow
derivative of cos is -sin --> dy/dx = -10sin x
anonymous
  • 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
  • dumbcow
slope = dy/dx evaluated at some x

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anonymous
  • anonymous
im not familiar with that method. would you mind demonstrating it please?
dumbcow
  • dumbcow
what method? its just the rule, derivative of sin is cos derivative of cos is -sin
anonymous
  • anonymous
cos/-sin is the solution?
dumbcow
  • 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
  • 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
  • dumbcow
i already gave you the derivative, now just plug in pi/3 for x to get slope
anonymous
  • anonymous
-10sin(pi/3) ok
dumbcow
  • dumbcow
yep...now use unit circle or calculator to find out what sin(pi/3) is
anonymous
  • anonymous
60degrees (1/2,sqrt3/2
anonymous
  • anonymous
what do i do with those numbers
dumbcow
  • dumbcow
which one refers to sin
anonymous
  • anonymous
sqrt3/2
dumbcow
  • dumbcow
cos() --> similar to x coord sin() --> similar to y coord good so slope is -5sqrt(3)
anonymous
  • anonymous
ok
anonymous
  • anonymous
the line tangent?
dumbcow
  • dumbcow
y = mx+b m = -5sqrt3 (x,y) = (pi/3,5) find b
anonymous
  • anonymous
wait, i didnt mention that my homework software was satisfied with -10sinsqrt(3) as the slope
dumbcow
  • dumbcow
you mean...-10sin(pi/3)
anonymous
  • anonymous
haha yeah
anonymous
  • anonymous
my homework software accepted that
dumbcow
  • 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
  • anonymous
ok thanks now i have to do slope and tangent line for m2
anonymous
  • anonymous
wait
dumbcow
  • dumbcow
m2?? haha, same function different point
anonymous
  • anonymous
my software didnt accept the previous answer
dumbcow
  • dumbcow
what did you put in? did you enter it as y=mx+b form
anonymous
  • anonymous
i entered it in this form 5 -(-10sin(pi/3)*pi/3)
dumbcow
  • dumbcow
well what does that number represent ?
dumbcow
  • dumbcow
i was trying to help you, i didn't give you the final answer
anonymous
  • anonymous
hahaha
anonymous
  • anonymous
5 -(-10sin(pi/3)*pi/3) i got 5+10sin(pi/3)pi/3
anonymous
  • anonymous
when i simplify
dumbcow
  • dumbcow
good
dumbcow
  • dumbcow
not done yet...what did you just solve for? what answer are they looking for..."equation of tangent line"
anonymous
  • anonymous
the line that runs diagonally and only touches a parabola on its sides
dumbcow
  • 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
  • anonymous
our slope is 10sin(pi/3) and y-int is 5 right?
dumbcow
  • dumbcow
no
dumbcow
  • dumbcow
y = mx+b b = y - mx m = -10sin(pi/3) (x,y) = (pi/3,5) b = 5 -(-10sin(pi/3)*pi/3)
anonymous
  • anonymous
i dont see how it isnt. you have m=-10sin(pi/3) and y=5
dumbcow
  • 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
  • anonymous
ohhhhh ok
anonymous
  • anonymous
im not sure how you solve for b here
dumbcow
  • dumbcow
b = y - mx b = 5 -(-10sin(pi/3)*pi/3)
anonymous
  • anonymous
omg
anonymous
  • anonymous
i didnt see the y-mx
anonymous
  • anonymous
i got 45
dumbcow
  • dumbcow
really? after multiplying by pi....or did you round it to 45
dumbcow
  • dumbcow
i get 14.069
anonymous
  • anonymous
my calc was in radians mode
dumbcow
  • dumbcow
yeah thats what you want you could probably leave it as: b= 5+10(pi/3)*sin(pi/3),
anonymous
  • anonymous
ahh i didnt distribute the negative outside the parantheisis
anonymous
  • anonymous
whats next
dumbcow
  • dumbcow
enter the equation of tangent line y = ?
dumbcow
  • 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
  • anonymous
ok thanks m2, do find the derivative of the derivative of m1
dumbcow
  • dumbcow
i dunno what they mean by m2
dumbcow
  • dumbcow
is their another function, or another point ?? does it look like dy^2/dx
anonymous
  • anonymous
The slope of the normal line is m2 The equation of the normal line is y
dumbcow
  • 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
  • dumbcow
this will be easier for perpendicular lines: m1*m2 = -1 m2 = -1/m1
dumbcow
  • dumbcow
Then find equation of normal line same way solve for y_intercept b = y - m2*x
dumbcow
  • dumbcow
http://www.wolframalpha.com/input/?i=y+%3D+10cos%28x%29+find+normal+line+at+x%3Dpi%2F3
anonymous
  • anonymous
what is that link for?
anonymous
  • anonymous
thanks for the help
dumbcow
  • dumbcow
your welcome link is to check your answer

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