# Tutor profile: Daniella S.

## Questions

### Subject: Calculus

Integrate the following function: $$ \int_0^2(\mathrm{e}^{x}+\frac{1}{x^2+1})\mathrm{d}x $$

$$ \int_0^2(\mathrm{e}^{x}+\frac{1}{x^2+1})\mathrm{d}x = (\mathrm{e}^{2}+tan^{-1}(2))-(\mathrm{e}^{0}+tan^{-1}(0))$$ $$ \int_0^2(\mathrm{e}^{x}+\frac{1}{x^2+1})\mathrm{d}x = \mathrm{e}^{2}+tan^{-1}(2)-1$$

### Subject: Physics

A traffic light with a mass of 15 kg is suspended between two wires attached to poles. The left wire is at a 30 degree angle with the horizontal and the right wire is at a 45 degree angle. Neglecting the masses of the wires, find the tension in each wire.

Step 1: Draw a free-body diagram. Step 2: $$F_{netx}=T_{2x}-T_{1x}$$ $$0=T_{2x}-T_{1x}$$ $$T_{1x}=T_{2x}$$ $$T_{1}cos(30)=T_{2}cos(45)$$ $$T_{2}=1.225T_{1}$$ Step 3: $$F_{nety}=T_{1y}+T_{2y}-F_g$$ $$F_g=T_{1y}+T_{2y}$$ $$T_{1}sin(30)+T_{2}sin(45)=F_g$$ Step 4: $$T_{1}(0.5)+(1.225T_{1})(0.707)=mg$$ $$1.366T_{1}=(15kg)(9.8m/s^2)$$ $$T_{1}=108N$$ Step 5: $$T_{2}=1.225T_{1}$$ $$T_{2}=1.225(108N)$$ $$T_{2}=132N$$ Step 5: Therefore, the tension in the wire on the left is 108 Newtons and the tension in the wire on the right is 132 Newtons.

### Subject: Differential Equations

Differentiate the following function: $$ y=\frac{cos(\frac{1}{x})}{tan(x)} $$

$$ \frac{dy}{dx}=\frac{sin(\frac{1}{x})tan(x)-x^2sec^2(x)cos(\frac{1}{x})}{x^2tan^2(x)} $$

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