A computer is purchased for $\$ 625$. Its value in dollars after $t$ years can be approximated by $V(t)=\frac{625}{(2 t+1)^{2}}$, where $0 \leq \mathrm{t} \leq 4$. (a) Find the value of the computer 2 years after purchase. (b) At time $t=2$ years, use a derivative to estimate how much the value will change within the next year. (a) The value of the computer 2 years after purchase is $\$ \square$. (b) At time $\mathrm{t}=2$ years, the value of the computer will by approximately $\square$ dollars within the next year. (Type a whole number.)

Step 1 ：Given the function \(V(t)=\frac{625}{(2 t+1)^{2}}\), where \(V(t)\) represents the value of the computer \(t\) years after purchase.

Step 2 ：For part (a), we need to find the value of the computer 2 years after purchase. This can be found by substituting \(t=2\) into the given function.

Step 3 ：Substituting \(t=2\) into the function, we get \(V(2)=\frac{625}{(2*2+1)^{2}} = 25\). So, the value of the computer 2 years after purchase is \(\boxed{25}\) dollars.

Step 4 ：For part (b), we need to find the rate of change of the value of the computer at \(t=2\) years. This can be found by first finding the derivative of the function \(V(t)\), and then substituting \(t=2\) into the derivative.

Step 5 ：The derivative of the function \(V(t)\) is \(V'(t) = -\frac{2500}{(2t + 1)^3}\).

Step 6 ：Substituting \(t=2\) into the derivative, we get \(V'(2) = -\frac{2500}{(2*2 + 1)^3} = -20\).

Step 7 ：So, at time \(t=2\) years, the value of the computer will decrease by approximately \(\boxed{20}\) dollars within the next year.