solved this problem?

Question

a> a hemispherical bowl of radius 10 cm contains waster to a depth of h cm. Find the radius of the surface of the water as a function of h

b> the water level drops at a rate of 0.1 cm per hour. at what rate is the radius of the water decreasing when the depth is 5 cm?

Answer 1

a) let R be the radius of the hemispheree, r be the variable radius, and given h is the depth of the hemispher.
Now, imagine a set of right triangles with legs h and r. Angle between the hypotenuse and the height of such triangles vary from pi/4 to pi/2.
Accordingly, r= h* tan(theta)
For example when r=R=10, h will also be equal to 10
Then, tan(theta) = 1 ; theta = 45 degree
b) In this part of problem it is difficult to find the rate of change of radius because it involves several variables like theta, time, radius etc.
Although I wold like to furnish hints to find the rate of change.
Given that h= (0.1)t. then dh/dt = 0.1
Also., r= t/10 * tan(theta); find the derivative w.r.t
Here the hurdle is d(th)/dt,
If is related to r problem will be easy.
Algebraically, we can find the radius of the surface of water at the depth h=5
ie. R^2 -(R-h)^2= r^2
then r= squrt(75)
therefore, tan(th)=squrt(75)/5
Hence, theta= 60 degree
Accordingly, r=8.66
Rate of change of r at h=5 is the limit of the difference quotient of the given function as h tends to 5 and as theta tends to 60.

Answer 2

a> this is difficult to answer without a picture.
Draw a triangle from [the edge of the water at the bowl] to [the centre of the top of the water] to[ the centre of the top of the bowl] (in the air)

If the water is of height h then the length of the line from [the centre of the top of the water] to [the centre of the top of the bowl ] is 10-h

the line from the [centre of the top of the bowl] to [the edge of the water at the bowl] will be 10 (the radius)

Then using trigonometry where the square of the hypotenuse is the sum of the squares of the other sides:

10*10 = (10-h)*(10-h) + r*r (where r is the radius we are looking for)

r*r = 100 - (100 -20h + h*h)
r*r = 20h-h*h
r = square root of (20h-h*h)

b>
r when depth is 5cm is r = sqr(20*5 - 5*5) = 8.660...
r when depth is 4.9cm is r = sqr(20*4.9 - 4.9*4.9) = 8.6017...

so the rate that the radius is decreasing is 0.058.. cm per hour.

(If you want to work this out properly then you will need to differentiate the equation)