# Tutor profile: Cameron N.

## Questions

### Subject: Physics (Newtonian Mechanics)

A $$10\,kg$$ mass is suspended by two ropes. Rope 1 hangs at $$45°$$ from vertical and rope 2 hangs at $$30°$$ from the vertical. Find the tension in each rope.

Since the mass is static, we know that $$\sum{F}=0$$. We can split this up into components as $$\sum{F_x}=0,\sum{F_y}=0$$. The only forces are gravity and the tensions in the ropes. We can use the given angles to split the rope tensions into $$x$$ and $$y$$ components, where $$F_x = F_T sin(\theta)$$ and $$F_y = F_T cos(\theta)$$. The force of gravity is $$F_g \approx 98.1\,N$$. Combining all of these into our component static equations gives us: $$\sum{F_x}=T_1 sin(45°) - T_2 sin(30°)=0$$ $$\sum{F_y}=T_1 cos(45°) + T_2 cos(30°) - 98.1=0$$ We can solve this using either substitution or a matrix solution, we find: $$T_1 = 50.8\,N$$ $$T_2 = 71.8\,N$$

### Subject: Pre-Calculus

Find all values of $$x$$ for which $$p(x)=x^3 - 3x^2 - 10 x > 0$$.

The first step for this problem is to factor the polynomial. Polynomials only cross the x-axis at single roots (and triple roots, quintuple roots, etc.). Since there is no constant term we can factor out an $$x$$ for $$x(x^2-3x-10)>0$$. We then plug into the quadratic formula for: $$x_1,x_2=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}=\frac{3 \pm \sqrt{3^2 -4*1*(-10)}}{2*1}=5,-2$$ This lets us finish factoring to get $$x(x+2)(x-5)>0$$, meaning we have the roots $$-2,0,5$$. We then test a value to get started. $$p(-3)=-24$$, so we know that $$p(x)\leq0$$ for $$x\leq-2$$. This alternates at each of our roots, so $$p(x)>0$$ for $$-2<x<0$$ and $$x>5$$.

### Subject: Calculus

A small cylindrical tank, $$10\,cm$$ in height and $$1\,cm$$ in radius, is being filled with water. Water flows into the tank through a hose at a rate of $$20\frac{cm^3}{s}$$. However, water is also leaking through a crack in the bottom of the tank at a rate of $$h \frac{cm^3}{s}$$, where $$h$$ is the height of the water in the tank. How long does it take for the tank to fill?

The volume of a cylinder is given by $$V=\pi r^2 h$$. We can differentiate this to get $$\frac{dV}{dh}=\pi r^2$$. Since the radius of our cylinder is $$1\,cm$$, we can simplify this to $$\frac{dV}{dh}=\pi$$ as long as we keep using centimeters. With $$20\frac{cm^3}{s}$$ flowing into the tank and $$h \frac{cm^3}{s}$$ flowing out, we have $$\frac{dV}{dt}=20-h\,\frac{cm^3}{s}$$. Since the equation is in terms of $$h$$ but our derivative is in terms of $$t$$, we can't integrate it as-is. First we need to use the chain rule along with our previous equation. We have $$\frac{dh}{dt}=\frac{\frac{dV}{dt}}{\frac{dV}{dh}}=\frac{20-h}{\pi}$$. We can solve this with some manipulation and a change of variables: $$\frac{dh}{dt}=\frac{20-h}{\pi},\,u=20-h$$ $$\frac{dh}{20-h}=\frac{dt}{\pi},\,du=-dh$$ $$\frac{-du}{u}=\frac{dt}{\pi}$$ $$-ln\left|u\right|=\frac{t}{\pi}+C$$ $$-ln\left|20-h\right|=\frac{t}{\pi}+C$$ Our next step is to get rid of $$C$$ using our initial condition, that when $$t=0,\,h=0$$. Plugging this in we get $$C=-ln\left|20\right|$$. Plugging this back into our original equation gives us: $$-ln\left|20-h\right|=\frac{t}{\pi}-ln\left|20\right|$$ $$ln\left|20\right|-ln\left|20-h\right|=\frac{t}{\pi}$$ $$ln\left|\frac{20}{20-h}\right|=\frac{t}{\pi}$$ $$t=\pi\,ln\left|\frac{20}{20-h}\right|$$ When the tank is full $$h=10\,cm$$, so $$t=\pi\,ln\left|\frac{20}{20-10}\right|\approx2.18\,s$$.

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