To find the equation of a tangent to a curve:
|To find the equation of a tangent to a curve: |To find the equation of a normal to a curve: |To find the coordinates of maximum/ minimum points on a curve: |
|find the derivative, [pic] |find the derivative [pic]; |differentiate to get [pic] |
|find the gradient, m, of the tangent by substituting in the x-ccordinate|Substitute in the x-coordinate of the point to find the value of the |solve the equation [pic] |
|of the point; |gradient there. |find the y-coordinates of the points |
|use one of the following formulae to get the equation of the tangent: |the gradient of the normal is [pic]. |determine whether the points are a maximum or minimum EITHER using the |
|EITHER y = mx + c |Use one of the following formulae to get the equation of the normal: |second derivative OR by considering the gradient either side of the |
|OR [pic] |EITHER y = mx + c |point. |
| |OR [pic] | |
|Example: | |Recall: A turning point is a maximum if [pic] |
|Find the equation of the tangent to the curve [pic] at the point where |Applications of Differentiation |A turning point is a minimum if [pic]. |
|the curve crosses the y-axis. | | |
| | | |
|Solution: | | |
|Expanding the brackets: [pic] | | |
|Differentiate: [pic] | | |
|The curve crosses the y-axis at the point (0, -2). | | |
|The gradient of the tangent at x = 0 is: [pic] | | |
| | | |
|To find the equation of the tangent: | | |
| | | |
|EITHER: y = mx + c | | |
|y = 3x + c | | |
|Substitute x = 0, y = -2: -2 = 3(0) + c i.e. c = -2. | | |
|So equation is y = 3x - 2 | | |
| | | |
|OR: [pic] where m = 3 | | |
|Substitute x = 0, y = -2: [pic] | | |
|i.e. y = 3x - 2 | | |
| |Example: |Example: |
| |Find the equation of the normal to the curve [pic] at the point where x |Find the coordinates of the stationary points on the curve [pic]. |
| |= 4. | |
| | |Solution: [pic] |
| |Solution: |At a stationary point, [pic]. |
| |The curve can be written as [pic] |Therefore, [pic] or [pic]. |
| |Therefore, [pic] |Factorising gives (x + 1)(x – 4) = 0. |
| |When x = 4, [pic] |Therefore x = -1 or x = 4. |
| |and [pic] | |
| |So the gradient of the normal is [pic]. |When x = -1, y = 28. When x = 4, y = -97. |
| | |The 2nd derivative is [pic]. |
| |To find the equation of the tangent: |When x = -1, [pic] so (-1, 28) is a MAX. |
| |y = mx + c [pic] y = 4x + c |When x = 4, [pic] so (4, -97) is a MIN. |
| |Substitute x = 4, y = 2: 2 = 4(4) + c i.e. c = -14. | |
| |So equation is y = 4x – 14. | |
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