Integrating an Expression with d(h³)/dt

In summary, the conversation discusses an expression for the variation in time of fluid volume in a cone-shaped tank. It is given by dV/dt = F0 - K*h^(1/2), where V is the volume, h is the variable height of the fluid, R is the radius of the tank, and H is the height of the tank. By a trigonometric relation, it is shown that V = (pi*R²/3*H²)*h³. The main problem in finding an integral for this expression is the term d(h³)/dt, which can be solved using the chain rule.
  • #1
Marcos Domingues
2
0

Homework Statement


[/B]
I have this expression: dV/dt = F0 - K*h^(1/2); it describes a variation in time of a fluid volume V in a cone-shaped tank of total volume H*pi*R²/3;

By a trigonometric relation we get V = (pi*R²/3*H²)*h³; since tan a = H/R = h/r
where: R = radius of the tank; H = height of the tank; h = variable height of the fluid; r = radius related to a specific height h

So, when we replace V, we got the general expression (pi*R²/3*H²) d(h³)/dt = F0 - K*h^(1/2);

This d(h³) term is the main problem on the integration. How do I integrate this expression?

Homework Equations



(pi*R²/3*H²) d(h³)/dt = F0 - K*h^(1/2)

The Attempt at a Solution



Maybe consider d(h³) = dh*dh*dh and solve it as multiple integrals?
 
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  • #2
Marcos Domingues said:
So, when we replace V, we got the general expression (pi*R²/3*H²) d(h³)/dt = F0 - K*h^(1/2);
No, this wrong. You need to use the chain rule, i.e.
$$\frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt}$$,
since ##V=V(h(t))##.
 
  • #3
eys_physics said:
No, this wrong. You need to use the chain rule, i.e.
$$\frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt}$$,
since ##V=V(h(t))##.

That's right. Thanks.
 

Related to Integrating an Expression with d(h³)/dt

1. What is the purpose of integrating an expression with d(h³)/dt?

The purpose of integrating an expression with d(h³)/dt is to find the rate of change of a function or quantity over time. This can help in understanding how a system or process is changing and can be used to make predictions or analyze data.

2. How do you integrate an expression with d(h³)/dt?

To integrate an expression with d(h³)/dt, you need to use the power rule for derivatives, which states that the derivative of xⁿ is n*x^(n-1). This can be applied to the expression to find the derivative, which can then be integrated using standard integration techniques such as u-substitution or integration by parts.

3. What are some real-world applications of integrating an expression with d(h³)/dt?

Integrating an expression with d(h³)/dt has many real-world applications, such as in physics, where it can be used to calculate the velocity, acceleration, or position of an object over time. It is also commonly used in economics, biology, and engineering to model and analyze various systems and processes.

4. Can integrating an expression with d(h³)/dt be used to solve differential equations?

Yes, integrating an expression with d(h³)/dt is a crucial step in solving differential equations. By integrating both sides of a differential equation, you can eliminate the derivative and find the original function. This is known as the inverse operation of differentiation.

5. Are there any limitations to integrating an expression with d(h³)/dt?

While integrating an expression with d(h³)/dt is a useful tool, it does have some limitations. It can only be used for functions that are continuous and differentiable. Additionally, some integrals may not have closed-form solutions and require numerical methods to be solved.

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