A 0.3 ml dose of a drug is injected into a patient steadily for 0.65 seconds. At the end of this time, the quantity, , of the drug in the body starts to decay exponentially at a continuous rate of 0.35 percent per second. Using formulas, express as a continuous function of time, , in seconds.

Use the compound interest formula: A=P(1+i)^t.

P is the initial amount of the drug, 0.3ml.
i is - 0.0035.
t is in seconds.

You'll get:
A=0.3(1-0.0035)^t.

Sub in any value on t to find out how many ml are left t seconds after injection.

The 0.65 second injection time does not seem to be relevant as the question clearly states that the exponential decay starts AFTER the injection is completed.

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xero099 xero099 · Answer 2 · 2021-10-01T18:08:20+02:00

The drug quantity as a function of time is [tex]y = 0.3\cdot e^{-\frac{t}{285.214} }[/tex].

This case in an example of exponential decrease, since the drug quantity decreases at a constant ratio per unit time. The exponential decay model is expressed by the following formula:

[tex]y = y_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (1)

Where:

[tex]y[/tex] - Current drug quantity, in milliliters.
[tex]y_{o}[/tex] - Initial drug quantity, in milliliters.
[tex]t[/tex] - Time, in seconds.
[tex]\tau[/tex] - Time constant, in seconds.

If we know that [tex]\frac{y}{y_{o}} = 0.9965[/tex] and [tex]t = 1\,s[/tex], then the time constant and the exponential decay model are:

[tex]\tau \ln\frac{y}{y_{o}} = -t[/tex]

[tex]\tau = -\frac{t}{\ln \frac{y}{y_{o}} }[/tex]

[tex]\tau = - \frac{1\,s}{\ln 0.9965}[/tex]

[tex]\tau \approx 285.214\,s[/tex]

Then, the exponential decay model for the drug is: ([tex]y_{o} = 0.3 \,mL[/tex])

[tex]y = 0.3\cdot e^{-\frac{t}{285.214} }[/tex]

The drug quantity as a function of time is [tex]y = 0.3\cdot e^{-\frac{t}{285.214} }[/tex].

We kindly invite to see this question on exponential functions: https://brainly.com/question/3127939