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RUDIKE [14]
3 years ago
13

The half-life of caffeine in a healthy adult is 4.8 hours. Jeremiah drinks 18 ounces of caffeinated

Mathematics
1 answer:
statuscvo [17]3 years ago
7 0

We want to see how long will take a healthy adult to reduce the caffeine in his body to a 60%. We will find that the answer is 3.55 hours.

We know that the half-life of caffeine is 4.8 hours, this means that for a given initial quantity of coffee A, after 4.8 hours that quantity reduces to A/2.

So we can define the proportion of coffee that Jeremiah has in his body as:

P(t) = 1*e^{k*t}

Such that:

P(4.8 h) = 0.5 = 1*e^{k*4.8}

Then, if we apply the natural logarithm we get:

Ln(0.5) = Ln(e^{k*4.8})

Ln(0.5) = k*4.8

Ln(0.5)/4.8 = k = -0.144

Then the equation is:

P(t) = 1*e^{-0.144*t}

Now we want to find the time such that the caffeine in his body is the 60% of what he drank that morning, then we must solve:

P(t) = 0.6 =  1*e^{-0.144*t}

Again, we use the natural logarithm:

Ln(0.6) = Ln(e^{-0.144*t})

Ln(0.6) = -0.144*t

Ln(0.6)/-0.144 = t = 3.55

So after 3.55 hours only the 60% of the coffee that he drank that morning will still be in his body.

If you want to learn more, you can read:

brainly.com/question/19599469

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Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

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\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

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and now we derivate by x:

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the mixed partial derivates are equal :)

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