Given x=Sqrt(3)sin(2t) and y=4cos^2(t), where 0<t<pi. Show that dy/dx = kSqrt(3)tan(2t).

Differentiating the equation for x with respect to t, we get: dx/dt=2Sqrt(3)cos(2t);Take the reciprocal of dx/dt to get dt/dx=1/[2Sqrt(3)cos(2t)]Using a trigonometric identity on the equation for y, we get: y=2[1+cos(2t)];Differentiating the equation for y with respect to t, we get: dy/dt=-4sin(2t);Multiply dy/dt and dt/dx gives: dy/dx=-2/3 Sqrt(3)tan(2t).From the question we are asked to find k.Therefore, k=-2/3

PC

Related Maths A Level answers

All answers ▸

The line AB has equation 3x + 5y = 7. What is the gradient of AB?


Two forces P and Q act on a particle. The force P has magnitude 7 N and acts due north. The resultant of P and Q is a force of magnitude 10 N acting in a direction with bearing 120°. Find the magnitude of Q and the bearing of Q.


Intergrate 8x^3 + 6x^(1/2) -5 with respect to x


curve C with parametric equations x = 4 tan(t), y=5*3^(1/2)*sin(2t). Point P lies on C with coordinates (4*3^(1/2), 15/2). Find the exact value of dy/dx at the point P.