If ct794te013t measures the concentration in ng ml of a drug...

If C(t) = 79.4te^(-0.13t) measures the concentration, in ng/ml, of a drug in a person's system t hours after the drug is administered, how quickly is the concentration changing 14 hours from now? It is changing at enter your response here units per hour. If C(t) = 50te^(-0.3t) measures the concentration, in ng/ml, of a drug in a person's system t hours after the drug is administered, what is the peak concentration of the drug? The peak concentration is enter your response here ng/ml. If C(t) = 55te^(-0.2t) measures the concentration, in ng/ml, of a drug in a person's system t hours after the drug is administered, what is the peak concentration of the drug? The peak concentration is enter your response here ng/ml.

Transcript

00:01 Hi, in this question for the first part we are given with c of t to be equals to 79 .4 t times e raised to power minus 0 .13 times.

00:13 This measure the concentration and now we need to solve for t is equals to 14 as the rate of change.

00:24 So for that we first find the derivative of ct that will be equals to as 79 .4 times in this if we apply the product rule derivative of this times this we get e raised to power minus 0 .13 t derivative of this will be 1 plus this into derivative of this will be minus 0 .13 e raised to power minus 0 .13 t and this will be further written as c dash t to be equals to 79 .4 times e raised to power minus 0 .13 t here we get minus 0 .13 t e raised to power minus 0 .13 t.

01:07 Now for t as 14 we get this to be equals to 79 .4 e raised to power minus 0 .13 times 14 minus 0 .13 times 14 times e raised to power minus 0 .13 times 14 and solving this further we get the final result for c dash 14 as this to be equals to minus 10 .35 is the units per hour.

01:38 Next we need to solve for the second part b where we are given with ct to be equals to 50 t e raised to power minus 0 .3 t.

01:48 Here we need to find the peak concentration as we know we can find the peak concentration response for the time period that we need to solve for and hence for the peak concentration that can be calculated when its derivative is equal to 0.

02:07 Now solving this further for its derivative and we get c dash t to be equals to 50 times e raised to power minus 0 .3 t minus 0 .3 times t times e raised to power minus 0 .3 t and we need to equate this to 0 and solving this further we get 50 e raised to power minus 0 .3 t times 1 minus 0 .3 t is equal to 0 and solving this we get 1 minus 0 .3 t is equal to 0 that will give t to be equals to 1 by 0 .3 or further solving we get t to be equals to 10 by 3.

02:48 Now substituting the value of t as 10 by 3 into the given concentration we get 50 times 10 by 3 times e raised to power minus 0 .3 times here we can write 1 by 0 .3 only where this and this will get cancelled and further solving we get 500 by 3 into e raised to power minus 1 so we can write 500 by 3 e as it is...

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