Applications of Derivatives
Rate of Change of Quantities
Grade 12
Question:
<p>Let <em>I</em> be the purchase value of an equipment and <em>V(t)</em> be the value after it has been used for <em>t</em> years. The value <em>V(t)</em> depreciates at a rate given by differential equation \(\dfrac{dV(t)}{dt} = -k(T - t)\), where \(k > 0\) is a constant and <em>T</em> is the total life in years of the equipment. Then the scrap value <em>V(T)</em> of the equipment is</p>
<p>\(I - \dfrac{kT^2}{2}\)</p>
<p>\(I - \dfrac{k(T-t)^2}{2}\)</p>
<p>\(e^{-kT}\)</p>
<p>\(T^2 - \dfrac{1}{k}\)</p>
Step-by-Step Solution
Key Concept: Integrate the depreciation rate differential equation from t=0 to t=T using the initial condition V(0)=I to find the scrap value V(T). The depreciation depends on remaining lifespan (T-t), not elapsed time.
<p><strong>Step 1:</strong> Write the given differential equation:</p><p>$$\frac{dV(t)}{dt} = -k(T-t)$$</p><p><strong>Step 2:</strong> Integrate both sides with respect to t:</p><p>$$V(t) = \int -k(T-t)\,dt = -k\left(Tt - \frac{t^2}{2}\right) + C$$</p><p>$$V(t) = -kTt + \frac{kt^2}{2} + C$$</p><p><strong>Step 3:</strong> Apply initial condition V(0) = I:</p><p>$$I = -k(0) + \frac{k(0)^2}{2} + C$$</p><p>$$C = I$$</p><p><strong>Step 4:</strong> Write the complete solution:</p><p>$$V(t) = -kTt + \frac{kt^2}{2} + I$$</p><p><strong>Step 5:</strong> Find scrap value at t = T:</p><p>$$V(T) = -kT(T) + \frac{k(T)^2}{2} + I$$</p><p>$$V(T) = -kT^2 + \frac{kT^2}{2} + I$$</p><p>$$V(T) = -\frac{kT^2}{2} + I$$</p><p>$$V(T) = I - \frac{kT^2}{2}$$</p><p>∴ Answer: <strong>A</strong></p>
Correct Answer: A