The Curve C shows parametric equations x = 4tant and y = 5((3)^1/2)(sin2t) , Point P is located at (4(3)^1/2, 15/2) Find dy/dx at P.

First I would find the value of t at Point P - I would equate the x equation to 4(3)^1/2 and the y equation to 15/2. This would give me (Px,Py). After this I would then find dy/dt, and dx/dx by differentiating the two equations with respect to t. We can then find dy/dx by multiplying dy/dt by dt/dx ( we obtain dt/dx by finding the reciprocal of dx/dt ). With this we have an equation for dy/dx , now all we have to is substitue the value of t we found in the beginning to obtain dy/dx.

Answered by Arjun B. Maths tutor

3714 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Work out the equation of the normal to the curve y = x^3 + 2x^2 - 5 at the point where x = -2. [5 marks]


A curve has equation y^3+2xy+x^2-5=0. Find dy/dx.


Find the stationary points of the curve f(x) =x^3 - 6x^2 + 9x + 1


A curve C has the equation y=5sin3x + 2cos3x, find the equation of the tangent to the curve at the point (0,2)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences