The population of a virus in a host can be modeled by the function \( A \) that satisfies the differential equation \( \frac{d A}{d t}=t-\frac{1}{5} A \), where \( A \) is measured in millions of virus cells and \( t \) is measured in days for \( 5 \leq t<10 \). At time \( t=5 \) days, there are 10 million cells of the virus in the host. (a) Write an equation for the line tangent to the graph of \( A \) at \( t=5 \). Use the tangent line to approximate the number of virus cells in the host, in millions, at time \( t=7 \) days.

To find the equation of the tangent line at \( t = 5 \), we first need the slope, which is given by the function \( \frac{dA}{dt} \). We can calculate this at \( t = 5 \) using the differential equation: \[ \frac{dA}{dt} = 5 - \frac{1}{5}A \] Substituting \( A = 10 \) (since there are 10 million cells at \( t = 5 \)): \[ \frac{dA}{dt}\Bigg|_{t=5} = 5 - \frac{1}{5}(10) = 5 - 2 = 3 \] Thus, the slope of the tangent line at \( (5, 10) \) is 3. The equation of the tangent line can be written using the point-slope form: \[ A - 10 = 3(t - 5) \] Simplifying this gives: \[ A = 3(t - 5) + 10 \] \[ A = 3t - 15 + 10 = 3t - 5 \] Now, to approximate the number of virus cells at \( t = 7 \): \[ A(7) = 3(7) - 5 = 21 - 5 = 16 \] So, we can approximate that there will be about 16 million virus cells in the host at \( t = 7 \) days.