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3 years ago

6

The radius of a right circular cylinder is decreasing at a rate of 2 inches per minute while the height is increasing at a rate

of 8 inches per minute. determine the rate of change of the volume when r = 5 and h = 7.

1 answer:

gizmo_the_mogwai [7]3 years ago

6 0

V(t) = πr^2h
V'(t) = π(2rr'h + r^2h')
V'(t) = π(2(5)(-2)(7) + 5^2(8)
V'(t) = π(-140 + 200)
V'(t) = 60π in^3

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Juice: 48 fluid ounces for $2.07; 32 fluid ounces for $1.64

Alla [95]

Answer:

Step-by-step explanation:

Start by changing the dollars into cents.

2.07 dollars = 2.07 * 100 = 207 cents

1.64 dollars = 1.64 * 100 = 164 cents.

How much does 1 ounce cost for each?

207 cents / 48 = 4.3125 cents / oz

164 / 32 = 5.125

The better buy is the 48 fluid oz jar.

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3 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
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Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

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Answer:

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Step-by-step explanation:

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