The gradient of a curve is given by dy/dx = 3 - x^2. The curve passes through the point (6,1). Find the equation of the curve.

Since we differentiate a function to find the gradient of a curve at any point, we need to reverse that to find the equation of the curve. We do this by integrating with respect to x:If you have a constant (a number without x), it becomes (constant)x. In this case, 3 becomes 3xThen, if you do have an x, you add one to the power and divide by the new power. So, here, -x^2 will become (-x^3)/3If you're given a point and told to find the equation of the curve, you have to find the constant, c. This is because when you a constant, it becomes zero. To do this, you substitute the coordinates into your integrated form: y = 3x - (x^3)/3 + c. This leads to 1 = 3(6) - (6^3)/3 + c. Solve for c and you'll get 55.So the equation of the curve is y = 3x - (x^3)/3 + 55.Never forget +c!!

Answered by Darya N. Maths tutor

9131 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The curve C has the equation (x^2)+4xy-8(y^2)+27=0. Find dy/dx in terms of x and y.


Find the gradient of the exponential curve y(x)=(9e^(7x))/(12e^(2x)) at x=2/5


Find the general solution to the differential equation '' (x^2 + 3x - 1) dy/dx = (2x + 3)y ''


Differentiate the following: 5x^3


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