Math Problem Statement

find the total work done in moving a parcticle in a force field given by F = 3x * y deg - 52 deg + 10xk along the x = 1 + t ^ 2 y = 24 ^ 2 z = f ^ 3 from 1/x = 1 + f = 2

Solution

To calculate the total work done in moving a particle through a force field along a given path, we can use the line integral of the force field along the path. The work WW is given by:

W=CFdrW = \int_C \mathbf{F} \cdot d\mathbf{r}

Where:

  • F=(Fx,Fy,Fz)=3xyi^52j^+10xk^\mathbf{F} = (F_x, F_y, F_z) = 3x \cdot y \hat{i} - 52 \hat{j} + 10x \hat{k} is the force field.
  • r(t)=(x(t),y(t),z(t))\mathbf{r}(t) = (x(t), y(t), z(t)) is the parameterized path along which the particle moves.
  • dr=drdtdtd\mathbf{r} = \frac{d\mathbf{r}}{dt} dt is the differential displacement along the path.

Given: x(t)=1+t2,y(t)=242,z(t)=f3x(t) = 1 + t^2, \quad y(t) = 24^2, \quad z(t) = f^3

First, let's compute the derivatives dxdt\frac{dx}{dt}, dydt\frac{dy}{dt}, and dzdt\frac{dz}{dt}:

dxdt=2t,dydt=0(since y(t)=constant),dzdt=3f2dfdt\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 0 \quad (\text{since } y(t) = \text{constant}), \quad \frac{dz}{dt} = 3f^2 \cdot \frac{df}{dt}

Now, the differential displacement drd\mathbf{r} is:

dr=(2ti^,0j^,3f2dfdtk^)dtd\mathbf{r} = \left(2t \hat{i}, 0 \hat{j}, 3f^2 \frac{df}{dt} \hat{k} \right) dt

Now, compute Fdr\mathbf{F} \cdot d\mathbf{r}:

Fdr=(3xy)2tdt+(52)0dt+10x3f2dfdtdt\mathbf{F} \cdot d\mathbf{r} = \left(3x \cdot y \right) \cdot 2t \, dt + (-52) \cdot 0 \, dt + 10x \cdot 3f^2 \frac{df}{dt} \, dt

Substitute x(t)=1+t2x(t) = 1 + t^2 and y=576y = 576 (since y=242y = 24^2):

Fdr=(3(1+t2)576)2tdt+10(1+t2)3f2dfdtdt\mathbf{F} \cdot d\mathbf{r} = \left(3(1+t^2) \cdot 576 \right) \cdot 2t \, dt + 10(1 + t^2) \cdot 3f^2 \frac{df}{dt} \, dt

Simplify the integrand:

Fdr=(3456t(1+t2)+30f2(1+t2)dfdt)dt\mathbf{F} \cdot d\mathbf{r} = \left(3456t(1+t^2) + 30f^2(1 + t^2) \frac{df}{dt} \right) dt

Now, integrate from the initial point to the final point. The limits for tt and ff are given by the conditions 1/x=11/x = 1 and f=2f = 2. Therefore, at the initial point:

  • t1=0t_1 = 0 (since x(0)=1+02=1x(0) = 1 + 0^2 = 1)
  • t2t_2 is not given, but can be solved if x=1+t2x = 1 + t^2, which depends on the value of t2t_2 when f=2f = 2.

Therefore, we can calculate the work WW by integrating:

W=t1t2(3456t(1+t2)+30f2(1+t2)dfdt)dtW = \int_{t_1}^{t_2} \left(3456t(1+t^2) + 30f^2(1 + t^2) \frac{df}{dt} \right) dt

Without specific bounds on tt, a general solution can be complex. The above setup, however, is how you would compute it if given specific bounds.

Would you like to explore more details or discuss any specific part of this? Here are some questions to consider:

  1. What is the exact value of f(t)f(t) to find t2t_2 from the given path equations?
  2. How do you solve for t2t_2 when the relationship between ff and tt is given?
  3. Can we assume specific limits for t1t_1 and t2t_2 based on f(0)=0f(0) = 0 and f=2f = 2?
  4. How do you approach this integral when f(t)f(t) is explicitly given?
  5. Would you like a numerical approximation if explicit values for limits are provided?

Tip: Always check the parametrization bounds carefully when dealing with line integrals to ensure all variables are correctly accounted for.

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Math Problem Analysis

Mathematical Concepts

Line integrals
Force fields
Parametric equations

Formulas

Work done by a force field

Theorems

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Suitable Grade Level

Advanced undergraduate