Unit 3 - Objective 1 - Tangent Lines and Normal Lines


Example 1

To find the equation of the tangent line to the curve at a given point, you need to find the 1st derivative of the function. The derivative is then evaluated at a given point which will be the slope of the tangent line. Then using the point-slope form of the equation , you can find the equation of the tangent line.

To find the equation of the normal line (perpendicular to the tangent line) to the curve at a given point, you also find the 1st derivative of the function. The derivative is then evaluated at a given point which will be the slope of the tangent line. You then take the negative reciprocal of this slope to find the slope of the normal line. Then using the point-slope form of the equation, you can find the equation of the normal line.

Try it!

Find the equation of the tangent line to the curve

y = 2x² - 3 at the point (2,5)

You now need to find the slope of the tangent line to the curve at any point which would be:

= 4x

You now need to find the slope of the tangent line to the curve at the point (2,5). Evaluate the 1st derivative.

= 4 (2) = 8

You can now use the point-slope equation to find the equation of the tangent line to the curve at the point (2,5).

Some other correct forms of the same equation are:

-8x + y = -11 or -8x + y + 11 = 0


Example 2

If you want to find the equation of the normal line, you would take the negative reciprocal of 8 which would be -1/8. This would be the slope of the normal line. You can now use the point-slope equation to find the equation of the normal line to the curve at the point (2,5).

Some other forms of the same equation are:
8y = -x + 42
x + 8y = 42
x + 8y - 42 = 0

Try it!

Find the equation of the tangent line and normal line to the curve.

5x² + 20y² = 40 at the point (2,1)

Find the 1st derivative (dy/dx).

10x + 40y = 0

40y = -10x

=

Now evaluate the derivative to find the slope of the tangent line at the point (2,1).

The equation of the tangent line to the curve at the point (2,1) would be:

y - (1) = - (x - 2)

y - 1 = -x +

y - 1 = -x + 1

y = -x + 2

Other forms are:

x + y = 2

x + 2y = 4

The slope of the normal line would be 2 so the equation of the normal line to the curve at the point (2,1) would be:

y - 1 = 2 (x - 2)

y - 1 = 2x - 4

y = 2x - 3

Other forms are:

-2x + y = -3

-2x + y + 3 = 0



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