The point A lies on the curve with equation y = x^(1/2). The tangent to this curve at A is parallel to the line 3y-2x=1. Find an equation of this tangent at A. (PP JUNE 2015 AQA)  

My exact explanation would depend on the students level of understanding. The following answer assumes a basic understanding of differentiation and equations of lines; a) Gradient of tangent is equal to gradient of 3y - 2x = 1 which can be found by rearranging the equation. (2/3)  b) We can find out where along the curve y=x1/2 we get a gradient of 2/3 (since the curve is not linear)  c) This gives us the X coordinate which we can sub into the intial curve to get the y coordinate of point A.  d) We now have the coordinates of a point on the curve and the gradient of a curve. We can therfore use the general equation of a line to work out the equation of the line. (y - y1 = M ( x - x1) ). 

SC
Answered by Sefret C. Maths tutor

7037 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Use the formula 5p + 2q = t to find the value of q when p = 4 and t = 24. 6


A-level: solve 8cos^2(x)+6sin(x)-6=3 for 0<x<2(pi)


Find the gradient at the point (0, ln 2) on the curve with equation e^2y = 5 − e^−x . [4]


Given that y={(x^2+4)(x−3)}/2x, find dy/dx in its simplest form.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences