Given y = 2sin(θ) and x = 3cos(θ) find dy/dx.

The function is defined parametrically so we usually approach these questions using chain rule.Recall that: dy/dθ * dθ/dx = dy/dx So we will need to differentiate each expression individually then multiply them together.Differentiating the first with respect to θ we get:(1)   dy/dθ = 2cos(θ) ,then the expression for x gives us: dx/dθ = -3sin(θ) , We can then remember that differentials behave as fractions so we can flip both sides to get:(2)  dθ/dx = -1/3sin(θ) . Remembering chain rule we can multiply (1)*(2) to get dy/dx: dy/dθ * dθ/dx = 2cos(θ) * -1/3sin(θ) --> dy/dx = -2cos(θ)/3sin(θ)

Answered by Jacob C. Maths tutor

3861 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do you find the minimum of the equation sin^2(x) + 4sin(x)?


Edexcel January 2007 - Question 4 (Rates and Differential Equations)


Find the equation of the normal of the curve xy-x^2+xlog(y)=4 at the point (2,1) in the form ax+by+c=0


y=e^2x-11e^x+24 Find the stationary point, nature of the stationary point, the x-intercepts and the y-intercept (calculator allowed)


We're here to help

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

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy