11. If a snowball melts so that its surface area decreases at a rate of 1 cm/min, find the rate at which the diameter decreases when the diameter is 10 cm.
Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 miles away.
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experiments to set up the devices.
Related Rates
d) Write an equation that relates the quantities.
e) Finish solving the problem.
![**Problem 11**: If a snowball melts so that its surface area decreases at a rate of 1 cm²/min, find the rate at which the diameter decreases when the diameter is 10 cm.
For an educational setting, this problem involves understanding the relationship between the surface area of a sphere and its diameter. To solve this, one would typically use calculus, particularly related rates.
1. **Sphere Surface Area Formula**: The surface area \( A \) of a sphere is given by \( A = 4\pi r^2 \).
2. **Diameter and Radius Relation**: The diameter \( d \) of a sphere is twice the radius \( r \), so \( r = \frac{d}{2} \).
Given that the rate of change of the surface area \( \frac{dA}{dt} \) is -1 cm²/min (since the area is decreasing) and we need to find \( \frac{dd}{dt} \) when the diameter \( d \) is 10 cm.
Using the chain rule, relate the rates:
\[ \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} \]
\[ \frac{dr}{dt} = \frac{dd}{dt} \times \frac{1}{2} \]
Substitute the expressions and solve for \( \frac{dd}{dt} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92319225-79a4-4652-99cb-1e6dc15af7cc%2F905aacc5-c17a-48e0-9b73-2ed16ded465a%2Fzxqerp_processed.png&w=3840&q=75)
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