Given x = cost y=sin2t, how do you find the dy/dx terms parameter t and find the values parameter t points dy/dx = 0?

Question

Evelyn Agard · Accepted Answer

#(dy)/(dx)=0# at #t=(2m+1)pi/4#

In parametric equations #x=x(t)# and #y=y(t)#, #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

As #y=sin2t#, #(dy)/(dt)=cos2txx2=2cos2t#

and as #x=cost#, #(dx)/(dt)=-sint#

Hence #(dy)/(dx)=(-2cos2t)/sint#

As #sint!=0#, when #t=npi#

#(dy)/(dx)=0#, when #cos2t=0# but #t!=npi# i.e. #2t=(2m+1)pi/2#

or #t=(2m+1)pi/4#, where #m# is an integer

But note that #x# and #y# both are sinusoidal functions and hence their domain is limited to #[-1,1]# and hence as #x=cost#, #(dy)/(dx)=0# at #x=+-1/sqrt2#

graph{2xsqrt(1-x^2) [-2.527, 2.473, -1.11, 1.39]}

Christopher Allers · Answer

undefined To find $ \frac{dy}{dx} $ in terms of the parameter $ t $, where $ x = \cos(t) $ and $ y = \sin^2(t) $, you need to use the chain rule. First, express $ y $ explicitly as a function of $ x $ by substituting $ x = \cos(t) $ into the equation for $ y $. Then, differentiate $ y $ with respect to $ x $ using the chain rule. Once you have $ \frac{dy}{dx} $ in terms of $ x $, rewrite $ x $ in terms of $ t $, and simplify the expression.

To find the values of $ t $ where $ \frac{dy}{dx} = 0 $, set the expression for $ \frac{dy}{dx} $ equal to zero and solve for $ t $. These values of $ t $ will be the parameter points where the slope of the tangent line to the curve is zero.