Q. 52 In Exercises 51鈥�54, assume tha... [FREE SOLUTION]

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Chapter 14: Q. 52 (page 1107)

In Exercises 51鈥�54, assume that a thin wire in space follows the given curve C and that the density of the wire at any point on C is given by $蚁 (x, y, z)$ . Find the mass of the wire by computing the line integral of the density function along the specified curve.
C is the straight line through (5,鈭�10) and (0, 2), and $蚁 (x, y) = e^{x + y}$ .

Short Answer

Expert verified

The line integral of the density function along the specified curve is $M (x, y) = \frac{13}{7} [e^{2} - e^{- 5}]$ .

Step by step solution

Step 1. Given Information

We have to find the mass of the wire by computing the line integral of the density function along the specified curve.

C is the straight line through (5,鈭�10) and (0, 2), and $蚁 (x, y) = e^{x + y}$ .

Step 2. The given function is ρ(x,y)=ex+y

The points are $(5, 鈭� 10) and (0, 2)$

We write the line segment as a vector function:

role="math" localid="1651221528131" $r = (5, 鈭� 10) + t (0 - 5, 2 - (- 10)) 0 鈮� t 鈮� 1, or in parametric form r = (5, 鈭� 10) + t (- 5, 2 + 10) r = (5, 鈭� 10) + t (- 5, 12) x = 5 - 5 t, y = - 10 + 12 t x' = - 5, y' = 12$

Step 3. Now the integral is M(x,y)=∫Cρ(x,y)ds

$M (x, y) = {鈭�}_{0}^{1} e^{5 - 5 t - 10 + 12 t} \sqrt{{(- 5)}^{2} + {(12)}^{2}} d t M (x, y) = {鈭�}_{0}^{1} e^{- 5 + 7 t} \sqrt{25 + 144} d t M (x, y) = {鈭�}_{0}^{1} e^{- 5 + 7 t} \sqrt{169} d t M (x, y) = {鈭�}_{0}^{1} e^{- 5 + 7 t} 路 13 d t L e t - 5 + 7 t = u 7 d t = d u d t = \frac{1}{7} d u M (x, y) = \frac{13}{7} {鈭�}_{0}^{1} e^{u} d u M (x, y) = \frac{13}{7} {[e^{u}]}_{0}^{1} M (x, y) = \frac{13}{7} {[e^{- 5 + 7 t}]}_{0}^{1}$

Step 4. Now putting the value.

$M (x, y) = \frac{13}{7} {[e^{- 5 + 7 t}]}_{0}^{1} M (x, y) = \frac{13}{7} [e^{- 5 + 7 路 1} - e^{- 5 + 7 路 0}] M (x, y) = \frac{13}{7} [e^{- 5 + 7} - e^{- 5 + 0}] M (x, y) = \frac{13}{7} [e^{2} - e^{- 5}]$

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Short Answer

Step by step solution

Step 1. Given Information

Step 2. The given function is ρ(x,y)=ex+y

Step 3. Now the integral is M(x,y)=∫Cρ(x,y)ds

Step 4. Now putting the value.

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