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Worked Examples · Example 17

Q.Nembutal, a sodium salt (sodium pentobarbital) acts as a sedative and has many applications. Suppose Nembutal is used to anesthetize a dog. The dog is anesthetized when its blood stream concentration contains at least 45mg of sodium pentobarbital per kg of the dog's body weight. If the rate of change of sodium pentobarbital say, xx in the body, is proportional to the amount of drug present in the body. Show that sodium pentobarbital is eliminated exponentially from the dog's blood stream given that its half-life is 5 hours. What single dose should be administered in order to anesthetize a 50 Kg dog for 1 hour?

Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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Because the elimination rate is proportional to the amount present, x=x0e−ktx=x_0e^{-kt}; the 55-hour half-life gives k=log⁡25≈0.1386k=\tfrac{\log 2}{5}\approx0.1386/hr, and requiring the concentration to still be 4545 mg/kg after 11 hour gives an initial dose of about 25852585 mg for a 5050 kg dog.

dxdt=−kx ⇒ x=x0e−kt\dfrac{dx}{dt}=-kx\ \Rightarrow\ x=x_0e^{-kt}, with half-life t1/2=log⁡2kt_{1/2}=\dfrac{\log 2}{k}.

  • xx = amount (concentration) of drug at time tt
  • x0x_0 = initial dose/concentration, kk = elimination constant per hour.

Show exponential elimination

  1. “Rate of change proportional to amount present” and the drug is being removed, so

dxdt=−kx,k>0.\frac{dx}{dt}=-kx,\qquad k>0.

  1. Separate variables and integrate:

∫dxx=−∫k dt ⇒ log⁡x=−kt+C.\int\frac{dx}{x}=-\int k\,dt\ \Rightarrow\ \log x=-kt+C.

  1. Exponentiate: x=eCe−kt=x0e−ktx=e^{C}e^{-kt}=x_0e^{-kt} where x0=eCx_0=e^{C} is the amount at t=0t=0. Hence elimination is exponential.

Find kk from the half-life

  1. Half-life =5=5 hr means x=x02x=\tfrac{x_0}{2} at t=5t=5:

12=e−5k ⇒ 5k=log⁡2 ⇒ k=log⁡25=0.13863 hr−1.\frac12=e^{-5k}\ \Rightarrow\ 5k=\log 2\ \Rightarrow\ k=\frac{\log 2}{5}=0.13863\ \text{hr}^{-1}.

Required single dose

  1. Work in concentration cc (mg per kg body weight): c(t)=c0e−ktc(t)=c_0e^{-kt}. The dog stays anaesthetized while c≥45c\ge 45; the concentration is lowest at the end of the hour, so we need c(1)=45c(1)=45: …

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