Question

(Need precise answer for C)​

Answer 1

Hi there!

a.

We can find the x-coordinate by deriving an expression for the line and arc, then setting the two equal.

We are given that the graph has the equation:

`y = x - (x^2)/(120)`

Differentiate to solve for the slope:

`y' = 1 - (x)/(60)`

This is the slope of the line at 'p'. We are given that the line passes through (-15, 0), so we can use the point-slope formula:

`y - y_1 = m(x - x_1)\\\\y - 0 = (1 - (x)/(60))(x - (-15))\\\\y = (1 - (x)/(60))(x + 15)`

Since this line intersects at 'p', we can set this equation along with the equation for the arc equal to each other to solve for 'x'.

`(1 - (x)/(60))(x + 15) = x - (x^2)/(120)\\`

Solve using a graphing utility.

`x = 30`

b.

We can write an equation for line 'l' by solving for the slope at the 'p' value.

`m = 1 - (30)/(60) = 1 - 0.5 = 0.5`

Now, use the point-slope formula with this y-value and the x-axis intersection coordinates.

`y - 0 = 0.5(x + 15)\\\\y = 0.5x + 7.5`

c.

We can plug x = 60 into both function equations to find the distance.

For the arc:

`f(60) = 60 - (60^2)/(120) = 30`

For the line:

`f(60) = .5(60) + 7.5 = 37.5`

Subtract to find the distance. (QR)

`37.5 - 30 = 7.5`