How do you find the equation of the tangent line #y=sinx# at #(pi/6, 1/2)#?
2 Answers
Explanation:
The slope of the tangent line to a function
Where
#y=sinx#
the derivative is given by
#dy/dx=cosx#
The slope of the tangent line to
#m=dy/dx|_(x=pi/6)=cos(pi/6)=sqrt3/2#
The slope of the tangent line is
#y-y_1=m(x-x_1)#
#y-1/2=sqrt3/2(x-pi/6)#
Differentiate y and evaluate
The equation of the tangent line would then be
The equation would be
Explanation:
Let the equation of the tangent line be
Hence the equation of the tangent line is
You can verify this answer visually too
graph{(y-sqrt(3)/2x-1/2+(sqrt(3)pi)/12)(y-sin(x))=0 [-1.259, 1.781, -0.477, 1.04]}
The reason the equation of a tangent line is as shown above is because in a linear function,
By definition, the gradient of a tangent line is equal to the slope of a curve at the point where the tangent line meets the curve.
Hence,