How do you evaluate the line integral, where c is the line segment from (0,8,4) to (6,7,7)?

Answer 1

Let #vec(c)(t)=(6t,-t+8,3t+4)# and compute #int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt#, where #vec(F)(vec(c)(t))\cdot vec(c)'(t)# is a dot product of two vectors.

The parametric equation #vec(c)(t)=(6t,-t+8,3t+4)# will parameterize the line segment #c# from #(0,8,4)# to #(6,7,7)# as #t# increases from #t=0# to #t=1#.
If #vec(F)(x,y,z)# is the vector field you want to integrate over this line segment, the way to calculate the line integral is to calculate #int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt#, where #vec(F)(vec(c)(t))\cdot vec(c)'(t)# is the dot product of #vec(F)(vec(c)(t))# with #vec(c)'(t)#.
For example, if #vec(F)(x,y,z)=(x+y,y^2+z,xyz)#, then #vec(F)(vec(c)(t))=vec(F)(6t,-t+8,3t+4)#
#=(5t+8,(-t+8)^2+3t+4,6t(-t+8)(3t+4))#
#=(5t+8,t^2-13t+68,-18t^3+120t^2+192t)#.

Hence,

#vec(F)(vec(c)(t))\cdot vec(c)'(t)#
#=(5t+8,t^2-13t+68,-18t^3+120t^2+192t)\cdot (6,-1,3)#
#=30t+48-t^2+13t-68-54t^3+360t^2+576t#
#=-54t^3+359t^2+619t-20#,

and

#int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt=int_{0}^[1}(-54t^3+359t^2+619t-20)\ dt#
#=-27/2 t^4+359/3 t^3+619/2 t^2-20t|_{t=0}^{t=1}#
#=-27/2+359/3+619/2-20=1187/3=395.666666...#
You might wonder whether you would get the same answer if you used a different parameterization, such as #vec(c)(t)=(3t,-1/2 t+8,3/2 t+4)# for #0\leq t\leq 2#. The answer is "yes", and you can do the calculation to check it.
If you want to prove this fact in the general case, you have to think about what happens when you "change variables" in the general case. The formula for changing variables in a general case can be written as #int_{a}^{b} f(g(x))g'(x)\ dx=int_{g(a)}^{g(b)} f(u)\ du#, where #u=g(x)# is the substitution, #du=g'(x)\ dx#, #u=g(a)# when #x=a#, and #u=g(b)# when #x=b#.
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Answer 2

To evaluate the line integral along the line segment from (0,8,4) to (6,7,7), denoted as (C), you first parameterize the line segment using a parameter (t). You can express the parameterization as:

[x(t) = 6t, \quad y(t) = 8 - t, \quad z(t) = 4 + 3t]

Next, you need to find the differential arc length, (ds), along the curve using the formula:

[ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} dt]

Then, you evaluate the line integral of a vector function (F(x, y, z)) along the curve (C) using the formula:

[\int_C F(x, y, z) \cdot ds = \int_a^b F(x(t), y(t), z(t)) \cdot \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} dt]

where (a) and (b) are the initial and final parameter values, respectively.

By substituting the parameterization and (ds) into the line integral formula, you can evaluate the line integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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