JEE MAIN - Mathematics (2011 - No. 10)
Let $$I$$ be the purchase value of an equipment and $$V(t)$$ be the value after it has been used for $$t$$ years. The value $$V(t)$$ depreciates at a rate given by differential equation $${{dv\left( t \right)} \over {dt}} = - k\left( {T - t} \right),$$ where $$k>0$$ is a constant and $$T$$ is the total life in years of the equipment. Then the scrap value $$V(T)$$ of the equipment is
$$I - {{k{T^2}} \over 2}$$
$$I - {{k{{\left( {T - t} \right)}^2}} \over 2}$$
$${e^{ - kT}}$$
$${T^2} - {1 \over k}$$
Explanation
$${{dV\left( t \right)} \over {dt}} = - k\left( {T - t} \right)$$
$$ \Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)} dt$$
$$V\left( t \right) = {{k{{\left( {T - t} \right)}^2}} \over 2} + c$$
$$V\left( 0 \right) = I \Rightarrow I = {{K{T^2}} \over 2} + C$$
$$ \Rightarrow C = I - {{K{T^2}} \over 2}$$
$$\therefore$$ $$\,\,\,\,\,V\left( T \right) = 0 + C = I - {{K{T^2}} \over 2}$$
$$ \Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)} dt$$
$$V\left( t \right) = {{k{{\left( {T - t} \right)}^2}} \over 2} + c$$
$$V\left( 0 \right) = I \Rightarrow I = {{K{T^2}} \over 2} + C$$
$$ \Rightarrow C = I - {{K{T^2}} \over 2}$$
$$\therefore$$ $$\,\,\,\,\,V\left( T \right) = 0 + C = I - {{K{T^2}} \over 2}$$
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