Question

To overcome an infection an anti-biotic is injected into David’s bloodstream. After the injection the anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t. If the initial injection was 8 ccs and 5 ccs remain after 4 hours, estimate how many ccs will remain after 5 hours?

Answer 1

Answer:

4.45 ccs will remain after 5 hours.

Step-by-step explanation:

The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.

(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)

(dQ/dt) = -kQ

(dQ/Q) = -kdt

 ∫ (dQ/Q) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.

We obtain

In (Q/Q₀) = -kt

(Q/Q₀) = e⁻ᵏᵗ

Q(t) = Q₀ e⁻ᵏᵗ

the initial injection was 8 ccs and 5 ccs remain after 4 hours

Q₀ = 8 ccs,

At t = 4 hours, Q = 5 ccs

5 = 8 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.625

-kt = In (0.625) = -0.47

-4k = 0.47

k = 0.1175 /hour

Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ

At t = 5 hours, Q = ?

Q = 8 e⁻⁰•¹¹⁷⁵ᵗ

0.1175 × 5 = 0.5875

Q = 8 e(^-0.5875)

Q = 4.45 ccs

Hope this Helps!!!