Given that a curve C is represented parametrically by x=t²/2 + 3t y=t²-2t how do you find the value of dy/dx and show that d²2y/dx²=8/(t+3)^3 and Show that C has only one stationary point and that this is a minimum?

Question

Charles Anctil · Accepted Answer

Has a local minimum at #x = 4, y = -1#

If
#{(x=t^2+3t),(y=t^2-2t):}#

them

#{(dx/(dt)=2t+3),(dy/(dt)=2t-2):}#

and

#dy/dx=dy/(dt)dt/dx = (2t-2)/(2t+3)#

Stationary points are those points that observe

#dy/(dx)=0# So, for #t=1# we have one such a point. #x=4, y=-1#

Qualification is done considering the signal of #(d^2y)/(dx^2)#. In this case we have

#d/(dx)(dy/(dx)) = (d^2y)/(dx^2) =1/(dx/(dt))^2d/(dt)(dy/(dt))+dy/(dx) d/(dt)(1/(dx/dt))=10/(3+2t)^3#

For #t = 1# we have # (d^2y)/(dx^2) = 10/5^3 > 0# so the stationary point is a relative minimum point. The quadratic after parameter ellimination is a parabola

#p(x,y) =x^2-10x - 15 y - 2 x y + y^2=0 #

Attached the conic plot.

Scarlett Allison · Answer

undefined To find dy/dx, first, express y in terms of x by eliminating the parameter t. Then, differentiate y with respect to x.

To show that d²y/dx² = 8/(t+3)^3, first find d²y/dx² by differentiating dy/dx with respect to x and express it in terms of t.

To prove that C has only one stationary point and that it is a minimum, find the critical points by setting dy/dx = 0 and determine the nature of the stationary point using the second derivative test.