Consider the set of parametric equations x=sin(t), y=sin(2t). What is the equation of the tangent line at the origin with positive slope?

Question

5818 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-11T13:07:33

#y=2x#

Slope of tangent at a point #(x_0,f(x_0))# on the curve #y=f(x)# is given by the value of the derivative #(df)/(dx)# at #(x_0,f(x_0))#.

Here we are given a parametric equation of the type #(x(t),y(t))#. Slope of such parametric equation is given by #((dy)/(dt))/((dx)/(dt))#

As #x=sint# and #y=sin2t#, #(dy)/(dt)=2cos2t# and #(dx)/(dt)=cost#

Now at origin i.e. #(0,0)#, we have #t=0#, as it makes both #x# and #y# equal to #0#, and hence slope of tangent is given by

#(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(2cos2t)/cost#

and hence value of slope at #t=0# is #2# (observe that slope is positive) and equation of tangent is #y=2x#

graph{(y-2xsqrt(1-x^2))(y-2x)=0 [-2.5, 2.5, -1.25, 1.25]}