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 \( x=5 \). 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 to the graph of \( A(t) \) at \( t = 5 \), we need the value of \( A(5) \) and the derivative \( A'(5) \). We already know that \( A(5) = 10 \) million cells. Next, we calculate \( A'(5) \) using the given differential equation: \[ A'(t) = t - \frac{1}{5} A \] Substituting \( t = 5 \) and \( A = 10 \): \[ A'(5) = 5 - \frac{1}{5} \times 10 = 5 - 2 = 3 \] Now we have the point \( (5, 10) \) and the slope \( 3 \). The equation of the tangent line can be written in point-slope form: \[ y - 10 = 3(t - 5) \] Simplifying this, we get: \[ y = 3t - 15 + 10 \implies y = 3t - 5 \] To approximate the number of virus cells in the host at \( t = 7 \): \[ y = 3(7) - 5 = 21 - 5 = 16 \] Thus, the tangent line approximates that at \( t = 7 \) days, there are approximately 16 million virus cells in the host.