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4 years ago

Work the following area application problem.

Mathematics

2 answers:

Nookie1986 [14]4 years ago

7 0

SInce you need 1.5 feet of overhang, add 3 feet to each axis dimension 1.5 on each side):

Minor Axis = 18 + 3 = 21 feet

Major axis = 25 + 3 = 28 feet

The area of an ellipse is found by multiplying half the minor axis by half the major axis by PI.

1/2 minor axis = 21 / 2 = 10.5

1/2 major axis = 28 / 2 = 14

Using 3.14 for PI

Area = 10.5 x 14 x 3.14 = 147 x 3.14 = 461.6 sq ft

dolphi86 [110]4 years ago

7 0

This problem is not as simple as it may appear at first. The area of an ellipse is ...

... A = πab

where a and b are the semi-axes.

Here, it looks like you're expected to choose these to be 1.5 ft longer than half the given axes, so the area of the pool cover is about ...

... A = π(9 ft + 1.5 ft)(12.5 ft +1.5 ft)

... A = 147π ft² ≈ 461.8 ft²

_____

However, adding 1.5 ft of material to an ellipse results in a shape that is not an ellipse, but is slightly larger than the ellipse with the dimensions used above. The area of that may be about 462.5 ft² (found numerically).

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Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Fi

Liono4ka [1.6K]

Answer:

a) $f (x,y,z)= xy^2\sin(z)$

b) $\int_C F \cdot dr =0$

Step-by-step explanation:

Recall that given a function f(x,y,z) then $\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$ . To find f, we will assume it exists and then we will find its form by integration.

First assume that F = $\nabla f$ . This implies that

$\frac{\partial f}{\partial x} = y^2\sin(z)$ if we integrate with respect to x we get that

$f(x,y,z) = xy^2\sin(z) + g(y,z)$ for some function g(y,z). If we take the derivative of this equation with respect to y, we get

$\frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}$

This must be equal to the second component of F. Then

$2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)$

This implies that $\frac{\partial g}{\partial y}=0$ , which means that g depends on z only. So $f(x,y,z) = xy^2\sin(z) + g(z)$

Taking the derivative with respect to z and making it equal to the third component of F, we get

$xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)$

which implies that $\frac{dg}{dz}=0$ which means that g(z) = K, where K is a constant. So

$f (x,y,z)= xy^2\sin(z)$

b) To evaluate $\int_C F \cdot dr$ we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.