Question

an oil spill in the ocean is rough circular in shape, with radius R(t) feet t minutes after spill begins. the radius is increasing at the rate
R(t)=21/(0.07t+5)
a) find an expression for the radius R(t), assuming hat R=0 when t=0
b) what is the area A=pi0R^2 of the spill after 1 hour?

Answer 1

Do you mean R'(t) for the rate?

Answer 2

yes

Answer 3

All right. To find a function for the radius, you have to integrate R'(t). Once you have that, part (b) is easy. Want me to walk through the integration?

Answer 4

yes please

Answer 5

OK,
R'(t)=21/(0.07t+5)
\[\int\limits_{0}^{x} 21/(.07t+5) dt=21 \int\limits_{0}^{x} 1/(.07t+5)dt\]

Integrating 1/(x) gives ln(x), but we have to integrate by u-substitution. u=(.07t+5) gives \[21\int\limits_{0}^{x} 1/(u) dt\], du/dt=.07 dt/dx. Substitute 1/.07 dt into the integral (the 1/.07 will factor out to the beginning, leaving [21/.07\int\limits_{0}^{x} 1/(u) du\] This integral gives ln(u), so filling back in we have \[R(t)=21/.07 \times \ln (.07t +5) + C\] Note that C will make the expression equal zero when t=0, so \[C=-(21/.07 \times \ln(5))\]
With that, you should be able to find (b) easily enough.