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2 years ago

I need help with this

Mathematics

1 answer:

Sholpan [36]2 years ago

7 0

Step-by-step explanation:

71.5

....................

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Lemur [1.5K]

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3 years ago

Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the

ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let $r$ be the radius, $d$ the diameter, and $V$ the volume of the spherical balloon.

We know $\frac{dV}{dt}=20 \:{in^3/s}$ and we want to find $\frac{dr}{dt}$

The volume of a spherical balloon is given by

$V=\frac{4}{3} \pi r^3$

Taking the derivative with respect of time of both sides gives

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

We now substitute the values we know and we solve for $\frac{dr}{dt}$ :

$d=2r\\\\r=\frac{d}{2}$

$r=\frac{12}{2}=6$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044$

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and $\frac{dr}{dt}$ is:

$\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025$

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

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3 years ago

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3 years ago

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3 years ago

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NeX [460]

Answer:

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Step-by-step explanation:

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