Answer to Question #334942 in Calculus for Quân Jason

Question #334942

When a particle is located a distance x meter from the origin, a force F( x )(newton) acts on it. Given F'(x) = 16.3.arctan(x - 8.5), F(8.5)=0. How much work is done in moving it from x=0 to x=8.5.


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Expert's answer
2022-05-02T09:34:52-0400

We remind that the formula for the work is: A=x1x2F(s)dsA=\int_{x_1}^{x_2}F(s)ds, where x1=0x_1=0, x2=8.5x_2=8.5.

F(x)=16.3arctan(x8.5)F'(x)=16.3\,\arctan(x-8.5). We receive that: F=16.3arctan(x8.5)dxF=16.3\int\arctan(x-8.5)dx. Make the change t=x8.5t=x-8.5. We get: F=16.3arctan(t)dt=16.3tarctan(t)16.3t1+t2dt=16.3tarctan(t)16.312d(1+t2)1+t2=16.3tarctan(t)16.32ln(1+t2)+CF=16.3\int\arctan(t)dt=16.3\,t\,\arctan(t)-16.3\int\,\frac{t}{1+t^2}dt=16.3\,t\,\arctan(t)-16.3\int\,\frac12\frac{d(1+t^2)}{1+t^2}=16.3\,t\,\arctan(t)-\frac{16.3}{2}ln(1+t^2)+C

We substitute F(8.5)=0F(8.5)=0 and receive: C=(138.55arctan(8.5)8.15ln(73.25))C=-(138.55\,\arctan(8.5)-8.15\,ln(73.25)).

Calculate integrals: tarctan(t)dt=t22arctan(t)t22(1+t2)dt=t22arctan(t)(1212(1+t2))dt=t2+12arctan(t)t2+C1\int\,t\,\arctan(t)dt=\frac{t^2}{2}\arctan(t)-\int\frac{t^2}{2(1+t^2)}dt=\frac{t^2}{2}\arctan(t)-\int\left(\frac12-\frac{1}{2(1+t^2)}\right)dt=\frac{t^2+1}{2}\arctan(t)-\frac{t}{2}+C_1

ln(1+t2)dt=tln(1+t2)2t2(1+t2)dt=tln(1+t2)(22(1+t2))dt=tln(1+t2)2t+2arctan(t)+C2\int\,ln(1+t^2)dt=t\,ln(1+t^2)-\int\frac{2t^2}{(1+t^2)}dt=t\,ln(1+t^2)-\int\left(2-\frac{2}{(1+t^2)}\right)dt=t\,ln(1+t^2)-2t+2\,\arctan(t)+C_2

We point out that after the change t=x8.5t=x-8.5 the boundary points become: t1=8.5t_1=-8.5 and t2=0.t_2=0.

Thus, we receive:

A=16.38.50(tarctan(t)12ln(1+t2))dt=16.3(t2+12arctan(t)t2)8.5016.32(tln(1+t2)2t+2arctan(t))8.50615.95A=16.3\int_{-8.5}^{0}\left(t\,\arctan(t)-\frac12ln(1+t^2)\right)dt=16.3\left(\frac{t^2+1}{2}\arctan(t)-\frac{t}{2}\right)|_{-8.5}^0-\frac{16.3}{2}\left(t\,ln(1+t^2)-2t+2\,\arctan(t)\right)|_{-8.5}^0\approx615.95


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