A curve has parametric equations: x=(t-1)^3 and y= 3t - 8/(t^2). Find dy/dx in terms of t. Then find the equation of the normal at the point on the curve where t=2.

dx/dt = 3(t-1)2dy/dt = 3 + 16t-3dy/dx=(dy/dt)(dt/dx) dy/dx = 3 + 16t-3 / 3(t-1)2
At t=2 dy/dx= (3 + 16/8) / 3 = 5/3 Gradient of the normal = -3/5with t=2 y-4=0x-1=0 y=mx + c y - 4 = -3/5(x-1) 3x +5y -23 = 0

JH
Answered by Jasmin H. Maths tutor

3308 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Let z=x+yi such that 16=5z - 3z*, What is z?


(C3) Show that 4csc^2(x) - cot^2(x) = k can be expressed as sec^2(x) = (k-1)/(k-4) where k != 4


Differentiate the function f(x) = x^2 * e^2x with respect to x


A circle with equation x^2+y^2-2x+8y-40=0. Find the circle centre and the radius


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