Question
f a snowball melts so that its surface area decreases at a rate of 9 cm2/min, find the rate at which the diameter decreases when the diameter is 8 cm.
Incorrect: Your answer is incorrect.
cm/min
Incorrect: Your answer is incorrect.
cm/min
Answers
GPT-4o mini
To solve this problem, let's define the relevant variables in the context of the snowball (which is a sphere):
1. Let \( r \) be the radius of the sphere.
2. The surface area \( S \) of a sphere is given by the formula:
\[
S = 4\pi r^2
\]
We know that the surface area is decreasing at a rate of \( \frac{dS}{dt} = -9 \, \text{cm}^2/\text{min} \). We want to find the rate at which the diameter \( d \) is decreasing when \( d = 8 \, \text{cm} \).
First, let's relate the diameter to the radius:
\[
d = 2r \implies r = \frac{d}{2}
\]
When the diameter is 8 cm, the radius is:
\[
r = \frac{8}{2} = 4 \, \text{cm}
\]
### Step 1: Find \( \frac{dr}{dt} \)
We can differentiate the surface area formula with respect to time \( t \):
\[
\frac{dS}{dt} = \frac{d(S)}{dr} \cdot \frac{dr}{dt}
\]
Calculating \( \frac{dS}{dr} \):
\[
\frac{dS}{dr} = 8\pi r
\]
Now substituting \( r = 4 \, \text{cm} \):
\[
\frac{dS}{dr} \bigg|_{r=4} = 8\pi (4) = 32\pi \, \text{cm}^2/\text{cm}
\]
Substituting into the rate of change formula:
\[
-9 = (32\pi) \cdot \frac{dr}{dt}
\]
Solving for \( \frac{dr}{dt} \):
\[
\frac{dr}{dt} = \frac{-9}{32\pi} \, \text{cm/min}
\]
### Step 2: Find \( \frac{dd}{dt} \)
Since the diameter \( d = 2r \), we differentiate \( d \) with respect to \( t \):
\[
\frac{dd}{dt} = 2\frac{dr}{dt}
\]
Substituting \( \frac{dr}{dt} \):
\[
\frac{dd}{dt} = 2 \left(\frac{-9}{32\pi}\right) = \frac{-18}{32\pi} = \frac{-9}{16\pi} \, \text{cm/min}
\]
Thus, at the moment when the diameter is 8 cm, the rate at which the diameter decreases is:
\[
\frac{dd}{dt} \approx -0.179 \, \text{cm/min} \quad (\text{approximately, if you want a numerical value})
\]
So the final answer for the rate at which the diameter decreases is:
\[
\frac{dd}{dt} = -\frac{9}{16\pi} \, \text{cm/min}
\]
This is the rate of decrease of the diameter when the diameter is 8 cm.
1. Let \( r \) be the radius of the sphere.
2. The surface area \( S \) of a sphere is given by the formula:
\[
S = 4\pi r^2
\]
We know that the surface area is decreasing at a rate of \( \frac{dS}{dt} = -9 \, \text{cm}^2/\text{min} \). We want to find the rate at which the diameter \( d \) is decreasing when \( d = 8 \, \text{cm} \).
First, let's relate the diameter to the radius:
\[
d = 2r \implies r = \frac{d}{2}
\]
When the diameter is 8 cm, the radius is:
\[
r = \frac{8}{2} = 4 \, \text{cm}
\]
### Step 1: Find \( \frac{dr}{dt} \)
We can differentiate the surface area formula with respect to time \( t \):
\[
\frac{dS}{dt} = \frac{d(S)}{dr} \cdot \frac{dr}{dt}
\]
Calculating \( \frac{dS}{dr} \):
\[
\frac{dS}{dr} = 8\pi r
\]
Now substituting \( r = 4 \, \text{cm} \):
\[
\frac{dS}{dr} \bigg|_{r=4} = 8\pi (4) = 32\pi \, \text{cm}^2/\text{cm}
\]
Substituting into the rate of change formula:
\[
-9 = (32\pi) \cdot \frac{dr}{dt}
\]
Solving for \( \frac{dr}{dt} \):
\[
\frac{dr}{dt} = \frac{-9}{32\pi} \, \text{cm/min}
\]
### Step 2: Find \( \frac{dd}{dt} \)
Since the diameter \( d = 2r \), we differentiate \( d \) with respect to \( t \):
\[
\frac{dd}{dt} = 2\frac{dr}{dt}
\]
Substituting \( \frac{dr}{dt} \):
\[
\frac{dd}{dt} = 2 \left(\frac{-9}{32\pi}\right) = \frac{-18}{32\pi} = \frac{-9}{16\pi} \, \text{cm/min}
\]
Thus, at the moment when the diameter is 8 cm, the rate at which the diameter decreases is:
\[
\frac{dd}{dt} \approx -0.179 \, \text{cm/min} \quad (\text{approximately, if you want a numerical value})
\]
So the final answer for the rate at which the diameter decreases is:
\[
\frac{dd}{dt} = -\frac{9}{16\pi} \, \text{cm/min}
\]
This is the rate of decrease of the diameter when the diameter is 8 cm.