A curve is defined by the parametric equations x=(t-1)^3, y=3t-8/(t^2), t is not equal to zero. Find dy/dx in terms of t.

We see that x and y are both expressed in terms of t and as we are looking to define dy/dx in terms of t, the first step we take is to find the derivatives of x and y in terms of t as follows.x = (t-1)^3dx/dt = 3(1)(t-1)(3-1) . (By the chain rule) = 3(t-1)2y = 3t - 8t(-2) (Note that y = 3t - 8t^(-2) to facilitate the use of chain rule)dy/dt = 3 - (8)(-2)t(-2-1) = 3 + 16t(-3)Now to find dy/dx, just divide dy/dt by dx/dt as follows.(dy/dt)/(dx/dt) = (3 + 16t(-3))/3(t-1)2 (the dt's cancel out)dy/dx = (3 + 16t(-3))/3(t-1)2 As Required.

Answered by James F. Maths tutor

5259 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I differentiate y=x^x?


Solve for x (where 0<x<360) 2sin^2(x) - sin(x) - 1 = 0


Prove that 2cot2x+tanx=cotx


What is the sum of the geometric series 1 + 1/3 + 1/9 + 1/27 ...


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences