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Andrei [34K]
3 years ago
15

Find the line integral with respect to arc length ∫C(8x+5y)ds, where C is the line segment in the xy-plane with endpoints P=(7,0

) and Q=(0,5). (a) Find a vector parametric equation r⃗ (t) for the line segment C so that points P and Q correspond to t=0 and t=1, respectively. r⃗ (t)= (b) Using the parametrization in part (a), the line integral with respect to arc length is ∫C(8x+5y)ds=∫ba dt with limits of integration a= and b= (c) Evaluate the line integral with respect to arc length in part (b). ∫C(8x+5y)ds=
Mathematics
1 answer:
puteri [66]3 years ago
4 0

a. Parameterize C by

\vec r(t)=(1-t)(7,0)+t(0,5)=(7-7t,5t)

with 0\le t\le1

b. The integral is to be computed over the range of the parameter t.

c. The integral has a value of

\displaystyle\int_C(8x+5y)\,\mathrm dS=\int_0^1(8x(t)+5y(t))\|\vec r(t)\|\,\mathrm dt

\displaystyle=\int_0^1(8(7-7t)+5(5t))\|(-7,5)\|\,\mathrm dt

\displaystyle=\sqrt{74}\int_0^1(56-31t)\,\mathrm dt=\boxed{81\sqrt{\dfrac{37}2}}

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The idea is to use the tangent line to f(x)=\sqrt x at x=25 in order to approximate f(25.3)=\sqrt{25.3}.

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