Please help me with this problem and please walk me through each step i am really confused.
Sand is falling from a rectangular box container whose base measures 40 inches by 20 inches at a constant rate of 300 cubic inches per minute.
a) how is the depth of the sand in the box changing?
b) the sand is form is forming a conical pile. At a particular moment, the pile is 23 inches high and the diameter of the base is 16 inches. The diameter of the base at this moment is increasing at 1.5 inches per minute, at this moment ,
1. how fast is the area of the circular base of the cone increasing?
2. How fast is the height of the pile increasing?
a) let the height of the sand in the box be h inches
V= (40)(20)(h) = 800h
dV/dt = 800 dh/dt
-300 = 800dh/dt
dh/dt = -300/800 = -3/8 inches per minute
b) part 2
V = (1/3)π r^2 h
dV/dt = (13)π r^2 dh/dt + (2/3)π r h dr/dt
plugging in our information ...
300 = (1/3)π(64)(dh/dt) + (2/3)π(8)(23)(.75)
solve for dy/dt
part 1
A = πr^2
r^2 = A/π
V = (1/3)π r^2 h
r^2 = 3V/(πh)
A/π = 3V/(πh)
A = 3V/h
dA/dt = (h(3dV/dt) - 3V dh/dt)/h^2
see where you can go from here
I am really confused on part 1. I don't even know what u did.
For part 2 where did u get .75 and did you mean to solve for dh/dt
the .75 comes from the fact that the diameter is increasing 1t 1.5in/min. The radius is half that.
and yes, he meant to solve for dh/dt, since that was the question asked.
Part 1 just uses the fact that the volume of a box is length*width*height.
can you say duh?
E=MC^2
a) To determine how the depth of the sand in the box is changing, we need to use the concept of rates.
Step 1: Let's start by finding the volume of the box using the formula:
Volume = length x width x height
Volume = (40 inches) x (20 inches) x (height of sand)
Step 2: Since the sand is falling at a constant rate of 300 cubic inches per minute, we can express the rate of change of the height of the sand as follows:
dV/dt = 300 cubic inches per minute, where V represents the volume and t represents time.
Step 3: To find how the height is changing, we differentiate the volume equation with respect to time (t), using the chain rule:
dV/dt = dV/dh x dh/dt, where h represents the height of the sand.
Step 4: Let's substitute the known values into the equation:
dV/dt = (40 inches) x (20 inches) x (dh/dt)
Step 5: Solving for dh/dt, we get:
dh/dt = (dV/dt) / (40 inches x 20 inches)
Step 6: Plug the given rate into the equation:
dh/dt = 300 cubic inches per minute / (40 inches x 20 inches)
Simplifying further:
dh/dt = 0.375 inches per minute
Therefore, the depth of the sand in the box is changing at a rate of 0.375 inches per minute.
b) Now let's solve the second part of the problem.
1. To find how fast the area of the circular base of the cone is increasing (dA/dt), we need to use the chain rule again.
Step 1: The formula for the area of a circle is:
A = πr^2, where r is the radius.
Step 2: We know that the diameter of the base is increasing at a rate of 1.5 inches per minute, so the rate of change of the radius (dr/dt) can be expressed as:
dr/dt = (1.5 inches per minute) / 2
Step 3: Differentiating the area equation with respect to time (t):
dA/dt = dA/dr x dr/dt
Step 4: Let's substitute the known values into the equation:
dA/dt = (dA/dr) x [(1.5 inches per minute) / 2]
Step 5: To find dA/dr, we differentiate the equation of the area of a circle with respect to the radius (r):
dA/dr = 2πr
Step 6: Plug the formula for dA/dr into the equation:
dA/dt = (2πr) x [(1.5 inches per minute) / 2]
Step 7: Apply the given values:
dA/dt = πr x (1.5 inches per minute)
Step 8: The diameter of the base is 16 inches at this particular moment, so the radius (r) is half of that:
r = 16 inches / 2 = 8 inches
Step 9: Plug the radius into the equation:
dA/dt = π(8 inches) x (1.5 inches per minute)
Simplifying further:
dA/dt = 12π inches² per minute
Therefore, the area of the circular base of the cone is increasing at a rate of 12π square inches per minute.
2. To find how fast the height of the pile is increasing (dh/dt), we can use a similar approach.
Step 1: We know that the diameter of the base is increasing at a rate of 1.5 inches per minute, so the rate of change of the radius (dr/dt) remains the same as before:
dr/dt = (1.5 inches per minute) / 2
Step 2: Instead of finding dA/dt, we want to find dh/dt. To do this, we need to relate the height (h) and the radius (r) of the cone.
Step 3: We can use the formula for the volume of a cone to relate h and r:
Volume = (1/3)πr^2h
Step 4: Since the height is changing, we can differentiate the volume equation with respect to time (t) using the chain rule:
dV/dt = (1/3)(2πrh)(dh/dt)
Step 5: Substitute the known values into the equation:
(300 cubic inches per minute) = (1/3)(2π(8 inches)(h))(dh/dt)
Step 6: Simplify and solve for dh/dt:
dh/dt = (3/2π(8 inches)(300 cubic inches per minute)) / h
Step 7: At this particular moment, the pile is 23 inches high, so substitute h = 23 inches:
dh/dt = (3/2π(8 inches)(300 cubic inches per minute)) / 23 inches
Simplifying further:
dh/dt = 25 cubic inches per minute
Therefore, the height of the pile is increasing at a rate of 25 cubic inches per minute.