Finding the Value of the Derivative of a Curve Using Differentiation

The video tutorial explains how to solve a calculus problem involving differentiation to find the maximum and minimum points of a curve. The equation y = X^3 - 27X + K is given, where K is a constant. The process involves differentiating the equation, solving for X values, and substituting these values back into the equation to find the Y coordinates B and D. The final task is to calculate the difference B - D, which is shown to be 108. The tutorial also outlines the marking scheme for the problem.

What is the equation of the curve discussed in the video?

• y = X^3 - 27X + K

• y = X^2 - 27X + K

• y = X^3 + 27X - K

• y = X^2 + 27X - K

What is the first step in finding the stationary points of the curve?

• Set DY/DX to 1

• Set Y to 0

• Set X to 0

• Set DY/DX to 0

After differentiating, what equation do we solve to find the X values?

• X^3 + 27 = 0

• X^3 - 27 = 0

• 3X^2 + 27 = 0

• 3X^2 - 27 = 0

What are the possible X values for the stationary points?

• X = 0 and X = 3

• X = 3 and X = -3

• X = 0 and X = -3

• X = 3 and X = 6

How do you calculate the value of B when A is -3?

• B = 27 - 81 + K

• B = -27 + 81 + K

• B = 27 + 81 - K

• B = -27 - 81 + K

What is the final calculated value of B - D?

• 54

• 108

• 81

• 27

Which step in the marking scheme involves substituting X values back into the original equation?

• Sixth mark

• First mark

• Second mark

• Fourth mark