A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a constant rate of 6 inches per second. When the radius is 4 feet, at what rate (in ft2/sec) is the total area A of the disturbed water changing?

dA/dt = 32π ft²/s

Step-by-step explanation:

r = the radius of the circle (in ft)

A = the area of the circle (in ft²)

t = time since the pebble hit the water (in s)

Rates of change;

dr/dt = 6 in/s = 6 x 0.0833 ft/s = 0.5 ft/s

Now, we are to find dA/dt when r = 4 ft

Area (A) = πr²

Let's Differentiate with respect to

t

Thus,

dA/dt = (d/dt) (πr²)

Using chain rule,

dA/dt = (2πr)(dr/dt)

Plugging in the relevant values to get ;

dA/dt = (2 x π x 4)/0.5 = 32π ft²/s

95.03 is the answer. hope this !

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).