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!!

DN

Related Maths A Level answers

All answers ▸

Find the tangent for the line y=x^3+3x^2+4x+2 at x=2


Express (x + 1)/((x^2)*(2x – 1)) in partial fractions


How should I go about solving a quadratic equation?


Use implicit differentiation to find the derivative of 2yx^2, with respect to x.