In the diagram below, the y-coordinate of point P is increasing at a rate of 4 units per minute while point Q remains in its fixed position. Find the rate of change of angle, in radians per minute, when y = 20.

y
^
| P (0, y)
|
|
|
|
|---|
|___|________ Q (4, 0)

Then, P (0 , 20). Can you calculate the slope since we have two coordinates given?

(0, 20) (4, 0)

m = (y2-y1)/(x2-x1)
m = (0 - 20) / (4 - 0)
m = 5

The slope is actually -5. And forget I think i understand this now.

Let the angle formed by the line by Ø

then
tanØ= y/4
4tanØ = y
dy/dt = 4sec^2 Ø (dØ/dt)

when y = 20
tanØ= 20/4 = 5
secØ = √26/1 = √26
sec^2Ø = 26

then
4 = 4(26)dØ/dt
dØ/dt = 1/26 rads/min

.00092

In order to find the rate of change of the angle when y = 20, we need to first find the angle between the x-axis and the line that connects points P (0, 20) and Q (4, 0). This can be done using trigonometry.

First, let's find the length of the line segment PQ.

The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Substituting the values of P(0, 20) and Q(4, 0) into the distance formula, we get:
d = sqrt((4 - 0)^2 + (0 - 20)^2)
d = sqrt(16 + 400)
d = sqrt(416)

Next, we can find the angle using the inverse tangent function:
angle = arctan((y2 - y1)/(x2 - x1))

Substituting the values of P(0, 20) and Q(4, 0) into the arctan function, we get:
angle = arctan((0 - 20)/(4 - 0))
angle = arctan(-20/4)
angle = arctan(-5)

Now, to find the rate of change of the angle with respect to time, we need to find the derivative of the angle equation. However, since the y-coordinate of point P is increasing at a rate of 4 units per minute, the x-coordinate remains constant at 0. Therefore, the rate of change of the angle will be zero since the angle does not change as y increases.

Hence, the rate of change of the angle, in radians per minute, when y = 20 is 0.