Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.

Answers

Answer 1

We have

x = t³   ===>   dx/dt = 3t²

y = t⁴   ===>   dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4[/tex]

[tex]\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}[/tex]

Answer 2

The line integral will be "[tex]\frac{265\sqrt{265} - 125 }{48}[/tex]". To understand the calculation, check below.

Line integral

According to the question,

Curves, x = t³ or,

            [tex]\frac{dx}{dt}[/tex] = 3t²

             y = t⁴ or,

            [tex]\frac{dx}{dt}[/tex] = 4t³  

Now, the line integral along C will be:

→ [tex]\int\limits_C {\frac{x}{y} } \, ds[/tex] = [tex]\int\limits^4_1 {\frac{t^3}{t^4} }[/tex] √(3t²)² + (4t³)² dt

             = [tex]\int\limits^4_1 {\frac{1}{t} }[/tex] √9t⁴ + 16t⁶ dt

             = [tex]\int\limits^4_1 {\frac{t^4}{t} }[/tex] √9 + 16t² dt

             = [tex]\int\limits^4_1 {\frac{t^2}{t} }[/tex] √9 + 16t² dt

             = [tex]\int\limits^4_1[/tex] t√9 + 16t² dt

             = [tex]\frac{1}{32}[/tex] [tex]\int\limits^4_1[/tex] √9 + 16t² d (9 + 16t²)

             = [tex]\frac{1}{32}\times \frac{2}{3}[/tex] [tex](9+16t^2)^{\frac{3}{2} } |_1^4[/tex]

             = [tex]\frac{1}{48}[/tex] ([tex]265^{\frac{3}{2}} - 25^{\frac{3}{2}[/tex])

             = [tex]\frac{265 \sqrt{265 } -125 }{48}[/tex]

Thus the above approach is correct.

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What amount of heat, in kj, would be involved in condensing 15. 8 g of ch₃oh? (∆hvap = 38. 0 kj/mol).

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Considering the definition of enthalpy of vaporization, the amount of heat that would be involved in condensing 15.8 g of CH₃OH is 18.8 kJ.

When a material changes phase from solid to liquid, or from liquid to gas, a certain amount of energy is involved in this phase change.

In case of phase change from liquid to gas, this amount of energy is known as enthalpy of vaporization (symbolized as ∆Hvap).

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In this case, you know that enthalpy of vaporization is ∆Hvap = 38.0 kJ/mol.

On the other side, you want to vaporize 15.8 g of CH₃OH.  Then, being the molar mass of the compound 32 [tex]\frac{g}{mole}[/tex] (that is, the amount of mass contained in one mole), the number of moles to be vaporized is:

[tex]15.8 gramsx\frac{1 mole}{32 grams} =0.49375 moles[/tex]

Then, the amount of heat that would be involved in condensing 15.8 g of CH₃OH is calculated taking into account the enthalpy of vaporization by:

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