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Alex787 [66]
3 years ago
8

Please help with this problem see image

Mathematics
1 answer:
babymother [125]3 years ago
5 0
Applying cosines law we have:
 5 ^ 2 = 7 ^ 2 + 6 ^ 2 - 2 * 7 * 6 * cos (F)
 Clearing the angle we have:
 cos (F) = (5 ^ 2 - 7 ^ 2 - 6 ^ 2) / (- 2 * 7 * 6)
 cos (F) = 0.714285714
 Then, clearing the angle:
 F = acos (0.714285714)
 F = 44.42 degrees
 Rounding:
 F = 44 degrees
 Answer:
 
F = 44 degrees
 
option 1
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balu736 [363]

Denote the circle of radius a by C. C is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of C (call it D):

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy

=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy

D is a circle of radius a, so we can write the double integral in polar coordinates as

\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta

a. For a=1, we have

\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}

b. Let I(a) denote the integral with unknown parameter a,

I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta

By the fundamental theorem of calculus,

I'(a)=12\pi(9a-a^3)

I(a) has critical points when

12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3

If a=0, then line integral is 0, so we ignore that critical point. For the other two, we would find I(\pm3)=243\pi.

8 0
3 years ago
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