# calculus

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please help i am lost at what is dy/dx and what is dy/dt. A trough is 10ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1ft. If the trough is being filled with water at a rate of 12ft^3/min, how fast is the water level rising when the water is 6 inches deep?

• calculus -

y and x are generic names of dependent and independent variables. In applications problems, we choose the names of variables to suit the situation, so that the problem can be better visualized.

In this prismatic tank problem, you can designate the variables as:
V = volume of water in the tank (cubic ft) (dependent variable)
h = height of water from the base (vertex of cross section) (independent variable and dependent variable)
t = time in minutes. (independent variable)

We would first calculate dV/dh (change of volume per unit rise of water level) and dV/dt is already given as 12 ft³/min.

The rate of rise of water level, dh/dt is simply (dV/dt)/(dV/dh).

Consider a horizontal slice of water of thickness dh at a height h above the bottom of the tank. The width at this point is 3'*h/1'=3h, the length of the tank is 10'. So the volume of the slice is 3h*10*dh. Therefore the volume increase per unit thickness is dV/dh = 30h.

Consequently the rise in level per unit time is
f(h)=
(dV/dt)/(dV/dh)
= 12 ft³/min. / 30h ft²
=(2/(5h)) ft/min. (h>0)

For h = 6" = 0.5', substitute h=0.5 in f(h) to get the rate.

• calculus -

When you see questions that contain the phrase "how fast" , they are usually answered with something like
12 feet/second, or 50 kmh, or 80 mph.
If y is some kind of distance then dy/dt is the rate at which y changes with respect to t, (the time).
If you have dy/dx it means, "the rate at which y changes in relation to x"
Slope is such an example. I the rise is y and the run is x, then dy/dx is the rate at which the rise (y) changes for every unit of the run (x)

for your problem above, let's first find an expression for the volume ...
Volume = area of triangle x 10

Make a sketch showing the depth of the water to be y ft and the length of the water level to be x ft
by ratio: x/y = 3/1
then y = x/3 or x = 3y

volume of water = (1/2)(x)(y)(10)
= 5xy = 5y(3y) = 15y^2

differentiate with respect to t (we need dy/dt)
dV/dt = 30y dy/dt
we are given dV/dt = 12 ft^3/min , and y = 6 inches = .5 ft

dV/dt = 15y dy/dt
12 = 15(.5) dy/dt
dy/dt = 12/7.5 = 1.6

So at the precise moment when the depth is 6 inches, the water level is rising at a rate of 1.6 ft/minute.

check my arithmetic.

• typo - calculus -

Sure enough I have a typo.

I had dV/dt = 30y dy/dt, but for some strange reason typed it as 15y dy/dt two lines later.

so near the end

dV/dt = 30y dy/dt
12 = 30(.5) dy/dt
dy/dt = 12/15 = 0.8 ft/min

So at the precise moment when the depth is 6 inches, the water level is rising at a rate of 0.8 ft/minute.