Question

Given y = sin(2x - π) + 1, find the (a) derivative, (b) equation of the tangent line at x = π/2, (c) equation of the normal line at x = π/2.

Answer 1

(a)
`y'= 2cos(2x - \pi)`

(b)
`y=2x - \pi + 1`

(c)
`y=-(x)/(2) +(\pi + 4)/(4)`

Explanation:

`y=sin(2x-\pi)+1`

Part (a)

Find the derivative of this function by using the chain rule and the power rule.

We know that the derivative of sinx = cosx. Find the derivative of this entire function first,
`sin(2x-\pi)+1`, then multiply this by the derivative of the inside function,
`2x-\pi`.

• 
  `(d)/(dx)(sin(2x-\pi)+1)`

Use the chain rule to find the derivative of sin(2x - π) + 1, which is cos(2x - π), then multiply this by the derivative of (2x - π). The derivative of π is 0, because it is a constant. The derivative of 2x is 2 based on the Power Rule.

• 
  `cos(2x-\pi) * 2`

Simplify this expression.

• 
  `2cos(2x - \pi)`

This is the derivative of
`y=sin(2x-\pi)+1`; therefore, we can write:

• 
  `y'= 2cos(2x - \pi)`

Part (b)

In order to find the equation of the tangent line at
`x=(\pi)/(2)`, we will need to find the slope of the tangent line and the x- and y- coordinates (we already know the x- cord).

The steps to finding the equation of the tangent line at a certain are:

1. Plug into y' to find the slope of the tangent line.
2. Plug into y to find the (x, y) coordinates.
3. Use point-slope to write our equation in slope-intercept form.

We know that y' = 2cos(2x - π). Let's plug x = π/2 into this equation for x to find the slope of the tangent line.

• 
  `y'((\pi)/(2) ) = 2cos(2((\pi)/(2))-\pi)`

Simplify inside the parentheses.

• 
  `y'((\pi)/(2) ) = 2cos((2\pi)/(2)-\pi)`
• 
  `y'((\pi)/(2) ) = 2cos(\pi - \pi)`
• 
  `y'((\pi)/(2) ) = 2cos(0)`
• 
  `y'((\pi)/(2) ) = 2`

Now we know that the slope of the tangent line is 2.

Let's plug x = π/2 into the original function, y.

• 
  `y((\pi)/(2))=sin(2((\pi)/(2))-\pi)+1`

Simplify inside the parentheses.

• 
  `y((\pi)/(2))=sin(0)+1`
• 
  `y((\pi)/(2))= 0+1`
• 
  `y((\pi)/(2))=1`

This tells us that the y-value, when x = π/2, equals 1. Our coordinates that we can use are (π/2, 1).

Now we can use point-slope form to write an equation for the tangent line to y at x = π/2.

Point-slope equation:

• 
  `y-y_1=m(x-x_1)`

We have
`(x_1, \ y_1)`, which are the x- and y- coordinates, and
`m`, which is the slope of the tangent line.

Substitute these values into the equation:

• 
  `y-(1)=2(x-((\pi)/(2)))`

Distribute 2 inside the parentheses.

• 
  `y-1=2x-(2 \pi)/(2)`

Add 1 to both sides of the equation.

• 
  `y=2x-(2\pi)/(2) + 1`
• 
  `y=2x - \pi + 1`

This is the equation of the tangent line of
`y=sin(2x-\pi)+1` at
`x=(\pi)/(2)`.

Part (c)

In order to find the equation of the normal line at x = π/2, we can use the information that the tangent line is perpendicular to the normal line.

This information is helpful because this means that their slopes are opposite reciprocals.

Let's use the point-slope equation again, but instead of m = 2, m will be the opposite reciprocal of 2 ⇒ -1/2. We will still use the same coordinate points.

• 
  `m=-(1)/(2) \ \ \ \ \ \ ((\pi)/(2), \ 1)`

• 
  `y-(1) = -(1)/(2)(x - ((\pi)/(2) ))`

Distribute -1/2 inside the parentheses.

• 
  `y-1=-(1)/(2)x + (\pi)/(4)`

Add 1 to both sides of the equation.

• 
  `y=-(1)/(2)x + (\pi)/(4)+1`
• 
  `y=-(1)/(2)x + (\pi)/(4) + (4)/(4)`
• 
  `y=-(1)/(2)x + (\pi+4)/(4)`

You can leave it written as this, or write it as:

• 
  `y=-(x)/(2) +(\pi + 4)/(4)`

This is the equation of the normal line of
`y=sin(2x-\pi)+1` at
`x=(\pi)/(2)`.