Math / Calculus

QuestionRemaining Time: 09:05:11
Air is being removed from a spherical balloon so that its volume decreases at a rate of $190 \mathrm{~cm}^{3} / \mathrm{s}$ . How fast is the radius of the balloon decreasing when the diameter is 10 cm ?
Do not include units in your answer. Your answer can be exact or approxinate, Ifitis approximate, round to three decimal places. 돈

Studdy Solution

STEP 1

What is this asking? How quickly is the balloon's radius shrinking at a specific moment? Watch out! Don't mix up diameter and radius!
Also, remember the rate of change is negative since the balloon is shrinking.

STEP 2

1. Set up the volume formula.
2. Differentiate with respect to time.
3. Plug in known values and solve.

STEP 3

We know the volume of a sphere is given by $V = \frac{4}{3} \cdot \pi \cdot r^3$ .
This formula connects the volume ( $V$ ) and the radius ( $r$ ).

STEP 4

We want to find how the rate of change of the volume relates to the rate of change of the radius.
We can do this by differentiating both sides of the volume formula with respect to time ( $t$ ).
Think of $V$ and $r$ as functions of time.

STEP 5

Using the chain rule, we get $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \cdot \pi \cdot r^3 \right) = \frac{4}{3} \cdot \pi \cdot 3 \cdot r^2 \cdot \frac{dr}{dt} = 4 \cdot \pi \cdot r^2 \cdot \frac{dr}{dt}.$ This equation now relates the rate of change of the volume ( $\frac{dV}{dt}$ ) to the rate of change of the radius ( $\frac{dr}{dt}$ ).

STEP 6

We're given that the volume is decreasing at a rate of $190 \text{ cm}^3/\text{s}$ , so $\frac{dV}{dt} = -190$ .
Notice the negative sign because the volume is decreasing.

STEP 7

We're also given that the diameter is $10$ cm at the moment we're interested in.
Since radius is half the diameter, the radius at that moment is $r = \frac{10}{2} = 5$ cm.

STEP 8

Now, let's plug these values into our differentiated equation: $-190 = 4 \cdot \pi \cdot (5)^2 \cdot \frac{dr}{dt}.$

STEP 9

Simplifying, we get $-190 = 100 \cdot \pi \cdot \frac{dr}{dt}.$

STEP 10

Now, we can isolate $\frac{dr}{dt}$ by dividing both sides by $100 \cdot \pi$ : $\frac{dr}{dt} = \frac{-190}{100 \cdot \pi} = \frac{-19}{10 \cdot \pi} = -\frac{19}{10\pi}.$

STEP 11

So, the radius is decreasing at a rate of $\frac{19}{10\pi}$ cm/s.
Approximating to three decimal places, we get $\frac{dr}{dt} \approx -0.605.$

STEP 12

The radius of the balloon is decreasing at a rate of approximately $-0.605$ .

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Studdy Solution

STEP 1

What is this asking? How quickly is the balloon's radius shrinking at a specific moment? Watch out! Don't mix up diameter and radius!
Also, remember the rate of change is negative since the balloon is shrinking.

STEP 2

1. Set up the volume formula.
2. Differentiate with respect to time.
3. Plug in known values and solve.

STEP 3

We know the volume of a sphere is given by $V = \frac{4}{3} \cdot \pi \cdot r^3$ .
This formula connects the volume ( $V$ ) and the radius ( $r$ ).

STEP 4

We want to find how the rate of change of the volume relates to the rate of change of the radius.
We can do this by differentiating both sides of the volume formula with respect to time ( $t$ ).
Think of $V$ and $r$ as functions of time.

STEP 5

STEP 6

We're given that the volume is decreasing at a rate of $190 \text{ cm}^3/\text{s}$ , so $\frac{dV}{dt} = -190$ .
Notice the negative sign because the volume is decreasing.

STEP 7

We're also given that the diameter is $10$ cm at the moment we're interested in.
Since radius is half the diameter, the radius at that moment is $r = \frac{10}{2} = 5$ cm.

STEP 8

Now, let's plug these values into our differentiated equation: $-190 = 4 \cdot \pi \cdot (5)^2 \cdot \frac{dr}{dt}.$

STEP 9

Simplifying, we get $-190 = 100 \cdot \pi \cdot \frac{dr}{dt}.$

STEP 10

Now, we can isolate $\frac{dr}{dt}$ by dividing both sides by $100 \cdot \pi$ : $\frac{dr}{dt} = \frac{-190}{100 \cdot \pi} = \frac{-19}{10 \cdot \pi} = -\frac{19}{10\pi}.$

STEP 11

So, the radius is decreasing at a rate of $\frac{19}{10\pi}$ cm/s.
Approximating to three decimal places, we get $\frac{dr}{dt} \approx -0.605.$

STEP 12

The radius of the balloon is decreasing at a rate of approximately $-0.605$ .

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Studdy Solution

STEP 1

What is this asking? How quickly is the balloon's radius shrinking at a specific moment? Watch out! Don't mix up diameter and radius!Also, remember the rate of change is negative since the balloon is shrinking.

STEP 2

1. Set up the volume formula.2. Differentiate with respect to time.3. Plug in known values and solve.

STEP 3

We know the **volume** of a sphere is given by V=43⋅π⋅r3V = \frac{4}{3} \cdot \pi \cdot r^3V=34​⋅π⋅r3.This formula connects the volume (VVV) and the radius (rrr).

STEP 4

We want to find how the **rate of change** of the volume relates to the **rate of change** of the radius.We can do this by differentiating both sides of the volume formula with respect to time (ttt).Think of VVV and rrr as functions of time.

STEP 5

STEP 6

We're given that the volume is decreasing at a rate of 190 cm3/s190 \text{ cm}^3/\text{s}190 cm3/s, so dVdt=−190\frac{dV}{dt} = -190dtdV​=−190.Notice the negative sign because the volume is *decreasing*.

STEP 7

We're also given that the diameter is 101010 cm at the moment we're interested in.Since radius is half the diameter, the radius at that moment is r=102=5r = \frac{10}{2} = 5r=210​=5 cm.

STEP 8

Now, let's plug these values into our differentiated equation: −190=4⋅π⋅(5)2⋅drdt.-190 = 4 \cdot \pi \cdot (5)^2 \cdot \frac{dr}{dt}.−190=4⋅π⋅(5)2⋅dtdr​.

STEP 9

Simplifying, we get −190=100⋅π⋅drdt.-190 = 100 \cdot \pi \cdot \frac{dr}{dt}.−190=100⋅π⋅dtdr​.

STEP 10

Now, we can **isolate** drdt\frac{dr}{dt}dtdr​ by dividing both sides by 100⋅π100 \cdot \pi100⋅π: drdt=−190100⋅π=−1910⋅π=−1910π.\frac{dr}{dt} = \frac{-190}{100 \cdot \pi} = \frac{-19}{10 \cdot \pi} = -\frac{19}{10\pi}.dtdr​=100⋅π−190​=10⋅π−19​=−10π19​.

STEP 11

So, the radius is decreasing at a rate of 1910π\frac{19}{10\pi}10π19​ cm/s.Approximating to three decimal places, we get drdt≈−0.605.\frac{dr}{dt} \approx -0.605.dtdr​≈−0.605.

STEP 12

The radius of the balloon is decreasing at a rate of approximately −0.605-0.605−0.605.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? How quickly is the balloon's radius shrinking at a specific moment? Watch out! Don't mix up diameter and radius!
Also, remember the rate of change is negative since the balloon is shrinking.

1. Set up the volume formula.
2. Differentiate with respect to time.
3. Plug in known values and solve.

We know the volume of a sphere is given by $V = \frac{4}{3} \cdot \pi \cdot r^3$ .
This formula connects the volume ( $V$ ) and the radius ( $r$ ).

We want to find how the rate of change of the volume relates to the rate of change of the radius.
We can do this by differentiating both sides of the volume formula with respect to time ( $t$ ).
Think of $V$ and $r$ as functions of time.

We're given that the volume is decreasing at a rate of $190 \text{ cm}^3/\text{s}$ , so $\frac{dV}{dt} = -190$ .
Notice the negative sign because the volume is decreasing.

We're also given that the diameter is $10$ cm at the moment we're interested in.
Since radius is half the diameter, the radius at that moment is $r = \frac{10}{2} = 5$ cm.

Now, let's plug these values into our differentiated equation: $-190 = 4 \cdot \pi \cdot (5)^2 \cdot \frac{dr}{dt}.$

Simplifying, we get $-190 = 100 \cdot \pi \cdot \frac{dr}{dt}.$

Now, we can isolate $\frac{dr}{dt}$ by dividing both sides by $100 \cdot \pi$ : $\frac{dr}{dt} = \frac{-190}{100 \cdot \pi} = \frac{-19}{10 \cdot \pi} = -\frac{19}{10\pi}.$

So, the radius is decreasing at a rate of $\frac{19}{10\pi}$ cm/s.
Approximating to three decimal places, we get $\frac{dr}{dt} \approx -0.605.$

The radius of the balloon is decreasing at a rate of approximately $-0.605$ .