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S_A_V [24]
2 years ago
13

EASY SCALE FACTOR SHOW WORK

Mathematics
1 answer:
Leni [432]2 years ago
6 0
The scale factor would be \frac{9.6}{6} =1.6

So the scale factor is 1.6
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Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2
exis [7]

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}

with determinant |J| = 12, hence the area element becomes

dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

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2 years ago
Analyze the diagram below and complete the instructions that follow.
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svet-max [94.6K]

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Sandy bought it for $81.66 [5/6×98]

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3 years ago
A model of a car is 4 inches long. If the actual car is 8 feet long, find the scale of the model. a. 4 in = 2 ft c. 8 in = 4 ft
Romashka-Z-Leto [24]

Answer:

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Step-by-step explanation:

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3 years ago
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Help me please <br><br><br><br>I couldn't solve it​
inysia [295]

\text{L.H.S}\\\\=\dfrac{\sin^4 A - \cos^4 A}{ \sin A + \cos A}\\\\\\=\dfrac{(\sin^2 A)^2-(\cos^2 A)^2}{\sin A + \cos A}\\\\\\=\dfrac{(\sin^2 A + \cos^2 A)(\sin^2 A-\cos^2 A)}{\sin A + \cos A}\\\\\\=\dfrac{1\cdot(\sin A+\cos A)(\sin A - \cos A)}{\sin A +\cos A}\\\\\\=\sin A - \cos A\\\\\\=\text{R.H.S}~~~ \\\\\text{Proved.}

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