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anonymous
 4 years ago
y=10cosx at the point (π/3,5), what is the derivative.
The slope of the tangent line is m1= .
The equation of the tangent line is y= .
The slope of the normal line is m2= .
The equation of the normal line is y
anonymous
 4 years ago
y=10cosx at the point (π/3,5), what is the derivative. The slope of the tangent line is m1= . The equation of the tangent line is y= . The slope of the normal line is m2= . The equation of the normal line is y

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precal
 4 years ago
Best ResponseYou've already chosen the best response.2the slope of the tangent line is y'(pi/3) The the derivative of your function and then sub (pi/3) into that equation then use \[yy_{1}=m(xx_{1})\] for your tangent line Take the slope that you found in part 1 and flip the fraction over and change its sign normal lines are perpendicular to the tangent lines so the slopes are opposite reciprocals. then use the same point but the new slope in this formula \[yy_{1}=m(xx_{1})\] That is your normal line
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