Answer :
To find the instantaneous rate at which the radius of the circular oil spill is growing after 29 minutes, we need to find the derivative of the function [tex]\( r(t) = \sqrt{9t} \)[/tex] with respect to [tex]\( t \)[/tex] and then evaluate it at [tex]\( t = 29 \)[/tex] minutes.
Step-by-step solution:
1. Rewrite the function for clarity:
The radius function is given by [tex]\( r(t) = \sqrt{9t} \)[/tex].
2. Express the function in a simpler form:
We can rewrite [tex]\( r(t) \)[/tex] as:
[tex]\[ r(t) = (9t)^{0.5} \][/tex]
3. Find the derivative [tex]\( r'(t) \)[/tex] using the chain rule:
According to the chain rule, the derivative of [tex]\( (9t)^{0.5} \)[/tex] is:
[tex]\[ r'(t) = 0.5 \cdot (9t)^{-0.5} \cdot 9 \][/tex]
Simplifying this, we get:
[tex]\[ r'(t) = \frac{0.5 \cdot 9}{(9t)^{0.5}} \][/tex]
4. Simplify the expression further:
[tex]\[ r'(t) = \frac{4.5}{(9t)^{0.5}} \][/tex]
Since [tex]\( (9t)^{0.5} \)[/tex] is the same as [tex]\( \sqrt{9t} \)[/tex], we can rewrite the derivative as:
[tex]\[ r'(t) = \frac{4.5}{\sqrt{9t}} \][/tex]
5. Evaluate the derivative at [tex]\( t = 29 \)[/tex]:
[tex]\[ r'(29) = \frac{4.5}{\sqrt{9 \cdot 29}} \][/tex]
6. Calculate the value step-by-step:
First, find [tex]\( 9 \cdot 29 \)[/tex]:
[tex]\[ 9 \cdot 29 = 261 \][/tex]
Next, find [tex]\( \sqrt{261} \)[/tex]:
[tex]\[ \sqrt{261} \approx 16.1555 \][/tex]
Then, calculate [tex]\( \frac{4.5}{16.1555} \)[/tex]:
[tex]\[ \frac{4.5}{16.1555} \approx 0.278543 \][/tex]
Therefore, the instantaneous rate at which the radius of the oil spill is growing after 29 minutes is approximately [tex]\(0.279 \, \frac{\text{ft}}{\min} \)[/tex].
Step-by-step solution:
1. Rewrite the function for clarity:
The radius function is given by [tex]\( r(t) = \sqrt{9t} \)[/tex].
2. Express the function in a simpler form:
We can rewrite [tex]\( r(t) \)[/tex] as:
[tex]\[ r(t) = (9t)^{0.5} \][/tex]
3. Find the derivative [tex]\( r'(t) \)[/tex] using the chain rule:
According to the chain rule, the derivative of [tex]\( (9t)^{0.5} \)[/tex] is:
[tex]\[ r'(t) = 0.5 \cdot (9t)^{-0.5} \cdot 9 \][/tex]
Simplifying this, we get:
[tex]\[ r'(t) = \frac{0.5 \cdot 9}{(9t)^{0.5}} \][/tex]
4. Simplify the expression further:
[tex]\[ r'(t) = \frac{4.5}{(9t)^{0.5}} \][/tex]
Since [tex]\( (9t)^{0.5} \)[/tex] is the same as [tex]\( \sqrt{9t} \)[/tex], we can rewrite the derivative as:
[tex]\[ r'(t) = \frac{4.5}{\sqrt{9t}} \][/tex]
5. Evaluate the derivative at [tex]\( t = 29 \)[/tex]:
[tex]\[ r'(29) = \frac{4.5}{\sqrt{9 \cdot 29}} \][/tex]
6. Calculate the value step-by-step:
First, find [tex]\( 9 \cdot 29 \)[/tex]:
[tex]\[ 9 \cdot 29 = 261 \][/tex]
Next, find [tex]\( \sqrt{261} \)[/tex]:
[tex]\[ \sqrt{261} \approx 16.1555 \][/tex]
Then, calculate [tex]\( \frac{4.5}{16.1555} \)[/tex]:
[tex]\[ \frac{4.5}{16.1555} \approx 0.278543 \][/tex]
Therefore, the instantaneous rate at which the radius of the oil spill is growing after 29 minutes is approximately [tex]\(0.279 \, \frac{\text{ft}}{\min} \)[/tex].
