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3 years ago

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A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 3.5 ft/s. (a) how rap

idly is the area enclosed by the ripple increasing when the radius is 5 feet?

1 answer:

MrRissso [65]3 years ago

5 0

(a) The area is given by

... A = π·r²

The rate of change of area is the derivative of this with respect to time.

... dA/dt = π·2r·dr/dt

For the given conditions, this evaluates to

... dA/dt = π·2·(5 ft)·(3.5 ft/s) = 35π ft²/s ≈ 109.96 ft²/s

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A is the right answer

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Step-by-step explanation:

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Find the slope of a line perpendicular to the line whose equation is 3y + 2x = 6

Charra [1.4K]

$3y + 2x = 6\\ \\the \ slope \ intercept \ form \ is : \\ \\ y= mx +b \\ \\3y=-2x+6 \ \ /:3\\ \\y=-\frac{2}{3}x+\frac{6}{3}\\ \\y=-\frac{2}{3}x +2 \\ \\ m = -\frac{2}{3} \\ \\ Answer : \ slope \ is \ m = -\frac{2}{3}$

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3 years ago

(X^2+y^2+x)dx+xydy=0<br> Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

<em>is</em> exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of <em>x</em>, and dividing this by $N(x,y)=xy$ gives an expression independent of <em>y</em>. If we assume $\mu=\mu(x)$ is a function of <em>x</em> alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by <em>x</em> on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution <em>F</em> satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to <em>x</em> gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to <em>y</em> gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

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3 years ago

The triangles are similar. Write a similarity statement for the triangles.

baherus [9]

Answer:

Option (2)

Step-by-step explanation:

In the two triangles ΔWVZ and ΔYXZ,

If the sides WV and XY are parallel and the segments WY and VX are the transverse.

∠X ≅ ∠V [Alternate angles]

∠W ≅ ∠Y [Alternate angles]

Therefore, ΔWVZ ~ ΔYXZ [By AA postulate of the similarity]

Option (2) will be the answer.

4 0

3 years ago