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Exercise 5 · Q10

Q.The space vehicles are supplied power from nuclear energy derived from radioactive isotopes. The output of the radioactive power supply for a certain satellite is given by the function, y=50e−0.004ty=50e^{-0.004t} where yy is in watts and tt is the time in days.

a) How much power will be available at the end of 90 days?
b) How long will it take for the amount of power to be half of its original strength?
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For y=50e−0.004ty=50e^{-0.004t}: at t=90t=90 days y≈34.88y\approx 34.88 W; and yy halves when t=log⁡20.004≈173.3t=\frac{\log 2}{0.004}\approx 173.3 days.

y=50e−0.004ty=50e^{-0.004t} — given power-output model, where yy = power in watts, tt = time in days; original strength =50=50 W at t=0t=0.

Part (a) — power after 90 days

  1. Substitute t=90t=90: y=50e−0.004×90=50e−0.36y=50e^{-0.004\times 90}=50e^{-0.36}.
  2. Evaluate e−0.36=0.697676e^{-0.36}=0.697676.
  3. y=50×0.697676=34.88 Wy=50\times 0.697676=34.88\ \text{W}.

Part (b) — time to fall to half strength

4. Half of the original 50 W is 25 W: 25=50e−0.004t⇒e−0.004t=1225=50e^{-0.004t}\Rightarrow e^{-0.004t}=\dfrac12.

5. Take logs: −0.004t=log⁡ ⁣(12)=−0.693147-0.004t=\log\!\left(\dfrac12\right)=-0.693147. …

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