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Kobotan [32]
1 year ago
15

to test an​ individual's use of a certain​ mineral, a researcher injects a small amount of a radioactive form of that mineral in

to the​ person's bloodstream. the mineral remaining in the bloodstream is measured each day for several days. suppose the amount of the mineral remaining in the bloodstream​ (in milligrams per cubic​ centimeter) t days after the initial injection is approximated by ​c(t) . find the rate of change of the mineral level with respect to time for days.
Mathematics
1 answer:
LUCKY_DIMON [66]1 year ago
3 0

Using derivatives, the rate of change of the mineral level with respect to time for 4 days is -0.012 mg/cm².

<h3>What is the missing information?</h3>

The function for the amount is missing, and is given by:

C(t) = \frac{1}{2}(6t + 1)^{-\frac{1}{2}}

The problems asks for the rate of change of the mineral level with respect to time for 4 days.

<h3>What is the rate of change after 4 days?</h3>

The rate of change after 4 days is given by:

C'(4).

Applying the chain rule, the derivative of C(t) is given as follows:

C^{\prime}(t) = \frac{1}{2}(6t+1)^\prime \times -\frac{1}{2} \times (6t+1)^{-\frac{3}{2}}

C^{\prime}(t) = -\frac{3}{2}(6t+1)^{-\frac{3}{2}}

When t = 4, we have that:

C^{\prime}(4) = -\frac{3}{2}(6(4)+1)^{-\frac{3}{2}}

C^{\prime}(4) = -\frac{3}{2}(25)^{-\frac{3}{2}}

Applying properties of exponents, we have that:

(25)^{-\frac{3}{2}} = \left(\frac{1}{25}\right)^{\frac{3}{2}} = \frac{1}{\sqrt{25^3}} = \frac{1}{25\sqrt{25}} = \frac{1}{125}

Hence:

C^{\prime}(4) = -\frac{3}{2} \times \frac{1}{125}

C'(4) = -0.012.

The rate of change of the mineral level with respect to time for 4 days is -0.012 mg/cm².

More can be learned about derivatives at brainly.com/question/23819325

#SPJ1

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Answer:

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Step-by-step explanation:

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