Question

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

Answer 1

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

What is the missing information?

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.

What is the rate of change after 4 days?

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.

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