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?
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.
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.
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.
To understand the concepts of dy/dx and dy/dt, let's start with some basic definitions.
In calculus, "dy/dx" represents the derivative, which measures the rate of change of the dependent variable y with respect to the independent variable x. It represents the instantaneous rate of change or the slope of a curve at a given point. Here, x and y are typically variables representing quantities that are related to each other.
On the other hand, "dy/dt" represents the derivative of the dependent variable y with respect to the independent variable t. This derivative measures how y is changing with respect to the variable t. The variable t is usually used to represent time in applications involving rates of change.
Now, let's apply these concepts to solve the specific problem you mentioned about the trough. We want to find how fast the water level is rising (rate of change) when the water is 6 inches deep.
Given information:
- The trough is 10 ft long.
- The ends of the trough have the shape of isosceles triangles with a base of 3 ft and a height of 1 ft.
- The trough is being filled with water at a rate of 12 ft^3/min.
To find the rate at which the water level is rising, we need to differentiate the formula for the volume of water in the trough with respect to time (dt), and then solve for dy/dt.
Let's break it down step by step:
Step 1: Express the volume of water in the trough as a function of the water depth y.
The trough has a triangular cross-section, so the volume of water in the trough can be expressed as the area of the cross-section multiplied by the length of the trough.
For a triangle, the formula for the area is (1/2) * base * height.
In this case, the base of the triangle is y (water depth), and the height is the corresponding height of the triangle at that water depth. Since we have isosceles triangles, the height can be determined by similar triangles.
Using similar triangles, we can set up the following proportion:
(base of triangle) / (height of triangle) = (base of large triangle) / (height of large triangle)
Substituting the given values, we have:
y / h = 3 / 1
Simplifying, we get:
y = 3h
Now, the volume of water in the trough can be expressed as:
V = (1/2) * y * h * L
where L is the length of the trough.
Substituting the given values, we have:
V = (1/2) * 3h * h * 10 = 15h^2
So, the volume of water in the trough is given by V = 15h^2.
Step 2: Differentiate the volume formula with respect to time (dt).
We are given that the trough is being filled with water at a rate of 12 ft^3/min. Therefore, the rate at which the volume is changing is given by dV/dt = 12.
Differentiating the volume formula with respect to time (dt), we get:
dV/dt = d(15h^2)/dt
Applying the Power Rule of differentiation, the derivative of h^2 is:
2h * (dh/dt)
So, we have:
12 = 2h * (dh/dt)
Step 3: Find dh/dt when the water depth y is 6 inches (or 0.5 ft).
We know that y = 3h. If y = 0.5 ft, then:
0.5 = 3h
Solving for h, we get:
h = 0.5 / 3 = 1/6 ft
Substituting this value of h into the equation from Step 2, we can find dh/dt:
12 = 2 * (1/6) * (dh/dt)
Simplifying, we get:
12 = (1/3) * (dh/dt)
Finally, solving for dh/dt, we have:
dh/dt = 12 * 3 = 36 ft/min
Therefore, the water level is rising at a rate of 36 ft/min when the water is 6 inches deep.
I hope this explanation helps clarify the concepts of dy/dx and dy/dt, as well as guide you through solving the problem.