The instantaneous rate at which the radius is growing after 31 minutes is [tex]0.449 \ units/min[/tex]
Explanation:
It is given that the radius of a circular oil spill after t minutes is given by
[tex]r(t)=\sqrt{25t}[/tex]
Simplifying, we have,
[tex]r(t)=5\sqrt{t}[/tex]
The instantaneous rate of change can be determine by differentiating with respect to t.
Thus, we have,
[tex]\frac{dr}{dt} =5\left(\frac{1}{2}(t)^{\frac{-1}{2}}\right)[/tex]
[tex]\frac{dr}{dt} =\left\frac{5}{2}(t)^{\frac{-1}{2}}\right[/tex]
[tex]\frac{dr}{dt} =\left\frac{5}{2\sqrt{t} }[/tex]
We need to determine the instantaneous rate of the radius after 31 minutes.
Hence, substituting [tex]t= 31[/tex] in the above expression, we get,
[tex]\frac{dr}{dt} =\left\frac{5}{2\sqrt{31} }[/tex]
Substituting the value of [tex]\sqrt{31}[/tex], we get,
[tex]\frac{dr}{dt} =\left\frac{5}{2(5.568)}[/tex]
Multiplying the denominator, we get,
[tex]\frac{dr}{dt} =\left\frac{5}{11.136}[/tex]
Dividing, we have,
[tex]\frac{dr}{dt} =0.449[/tex]
Thus, the instantaneous rate at which the radius is growing after 31 minutes is [tex]0.449 \ units/min[/tex]
The instantaneous rate at which the radius is growing after 31 minutes is 0.449
Given the radius of a circular oil spill after t minutes is given by r(t) = √25 t
The differential of the function with respect to t is the instantaneous velocity.
On differentiating:
[tex]r(t) = \sqrt{25t}\\r(t)=5\sqrt{t}\\r'(t)=5(\frac{1}{2\sqrt{t}} )[/tex]
If the radius is growing after 31 minutes, then:
[tex]r'(t)=5(\frac{1}{2\sqrt{t}} )\\r'(31)=5(\frac{1}{2\sqrt{31}} )\\r'(31) = 0.449[/tex]
Hence the instantaneous rate at which the radius is growing after 31 minutes is 0.449
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