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 ▸Integrate 5cos(3x - 1) with respect to x
Find the area enclosed by the curve y = cos(x) * e^x and the x-axis on the interval (-pi/2, pi/2)
What is a good method to go about sketching a polynomial?
Express (5sqrt(3)-6)/(2sqrt(3)+3) in the form m+nsqrt(3) where m and n are integers. [Core 1]