Question

the area of a blot of ink is growing such that t second A=3r²+t/5+7. Calculate the rate of increase of area after 5seconds​

Answer 1

Answer:

Explanation:

Given, Area of ink drop at time t = A = (3t^2 + t/5 + 7).

So, Rate of change in area at time t = dA/dt = (6t + 1/5 + 0) = (6t + 1/5) = {(30t + 1) / 5}.

Therefore, dA/dt at t = 5 seconds

= {(30 * 5 + 1) / 5} sq units per second

= (151 / 5) sq units per second

= 30.2 sq units per second.

Since, the magnitude of dA/dt is positive at t = 5 seconds; so the area of ink drop is increasing at that point of time and at the rate of 30.2 sq units per second.