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derivative of cos is -sin --> dy/dx = -10sin x
how will i find these two also The slope of the tangent line is m1= . The equation of the tangent line is y= .
slope = dy/dx evaluated at some x
im not familiar with that method. would you mind demonstrating it please?
what method? its just the rule, derivative of sin is cos derivative of cos is -sin
cos/-sin is the solution?
?? 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
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?
i already gave you the derivative, now just plug in pi/3 for x to get slope
yep...now use unit circle or calculator to find out what sin(pi/3) is
what do i do with those numbers
which one refers to sin
cos() --> similar to x coord sin() --> similar to y coord good so slope is -5sqrt(3)
the line tangent?
y = mx+b m = -5sqrt3 (x,y) = (pi/3,5) find b
wait, i didnt mention that my homework software was satisfied with -10sinsqrt(3) as the slope
my homework software accepted that
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)
ok thanks now i have to do slope and tangent line for m2
m2?? haha, same function different point
my software didnt accept the previous answer
what did you put in? did you enter it as y=mx+b form
i entered it in this form 5 -(-10sin(pi/3)*pi/3)
well what does that number represent ?
i was trying to help you, i didn't give you the final answer
5 -(-10sin(pi/3)*pi/3) i got 5+10sin(pi/3)pi/3
when i simplify
not done yet...what did you just solve for? what answer are they looking for..."equation of tangent line"
the line that runs diagonally and only touches a parabola on its sides
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
our slope is 10sin(pi/3) and y-int is 5 right?
y = mx+b b = y - mx m = -10sin(pi/3) (x,y) = (pi/3,5) b = 5 -(-10sin(pi/3)*pi/3)
i dont see how it isnt. you have m=-10sin(pi/3) and y=5
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
im not sure how you solve for b here
b = y - mx b = 5 -(-10sin(pi/3)*pi/3)
i didnt see the y-mx
i got 45
really? after multiplying by pi....or did you round it to 45
i get 14.069
my calc was in radians mode
yeah thats what you want you could probably leave it as: b= 5+10(pi/3)*sin(pi/3),
ahh i didnt distribute the negative outside the parantheisis
enter the equation of tangent line y = ?
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
ok thanks m2, do find the derivative of the derivative of m1
i dunno what they mean by m2
is their another function, or another point ?? does it look like dy^2/dx
The slope of the normal line is m2 The equation of the normal line is y
oh ok, so m1 refers to tangent line m2 refers to normal line which is perpendicular to tangent line at point (pi/3, 5)
this will be easier for perpendicular lines: m1*m2 = -1 m2 = -1/m1
Then find equation of normal line same way solve for y_intercept b = y - m2*x
what is that link for?
thanks for the help
your welcome link is to check your answer