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3 years ago

Find dy/dx. x^2y+xy^2=6

1 answer:

3 0

We are asked to evaluate dx/dy of the function x^2y+xy^2=6
we use implicit differentiation here:
2xy dx + 2xy dy = 0we can cancel 2xy from both terms in the left-hand side such that what is left is dx/dy = 0

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6c + 7c +7c + 8c + 9c nesesito de su ayuda plis

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Answer:

37c

Step-by-step explanation:

6c+(7c+7c)+8c+9c

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3 years ago

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Answer: 0.1 I might be wrong tho

Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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2 years ago

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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

-11-2+3(4*6^2/(-9))]

Lelechka [254]

Answer:

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Step-by-step explanation:

get a calculator and type it

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3 years ago

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