# Tutor profile: Darryl W.

## Questions

### Subject: Physics (Newtonian Mechanics)

A ball of 150g is thrown from $$h=0$$ with an initial velocity of 40 m/s upwards. Ignoring air resistance, Find: a) time in the air b) max height c) kinetic energy when it returns to $$h=0$$

for simplicity, we're going to use $$g = 10 m/s^2$$. We can also take a few shortcuts thanks to the symmetry of parabolic trajectories [of which 'directly upwards' is a special/degenerate case with $$sin(90) = 1$$] a) given $$v_i =40$$ and $$g = 10$$, the time of flight is $(t = \frac{2v_isin(\theta)}{g}=\frac{80}{10}=8s$) b) $(h_{max} = \frac{v_i^2}{2g} = \frac{40^2}{20} = 80$) c) $(KE = \frac{1}{2}mv^2$) Knowing that the ball will move upwards for 4s, then down for four more seconds, we can see that the velocity when it returns to $$h=0$$ has the same magnitude as it's initial velocity, $$40m/s$$. $(\frac{1}{2}*.15*40^2=120J$)

### Subject: Pre-Calculus

If $$(1)\;x^2+4y^2=40$$ and $$(2)\;xy=6$$, what will be the value of $$(3)\;x+2y$$

We have two equations and two unknowns, so this seems solvable, in general. However, rather than forcing it through the most naive method (something like, divide (2) by y, plug that value of x into (1), etc), a slightly more elegant and convenient way is available to us, if we notice that (3), x + 2y, is the square root of the first term and the square root of the second term of (1). With that as a clue, let’s try squaring (3) to get $((x+2y)^2=x^2+4xy+4y^2=z^2$) Where z is a stand in for the desired value. Conveniently, we have an equation for the value of the ‘xy’ term. $(x^2+4(6)+4y^2=z^2$) $(x^2+4y^2=z^2-24$) Thanks to (1), we also know the value of $$x^2+4y^2$$ explicitly $(x^2+4y^2=40=z^2-24$) $(40+24=z^2$) $(64=z^2$) Giving us a value for z (which was a stand in for the value requested) of $(\pm8=z$) The plus-minus deals with the ambiguity of the square root (we can confirm/double check our result with the other method, getting values of $$x,y=(2, 3)$$, and seeing that they match up to the parameters given in the question)

### Subject: Calculus

Differentiate $$2tan^2(x)-sec^2(x)$$

We could use trigonometric identities to simplify this expression a little first, but it’s not particularly necessary. We can solve it directly by using the chain rule and trigonometric derivatives $$\frac{d}{dx}tan(x)=sec^2(x)$$ and $$\frac{d}{dx}sec(x)=tan(x)sec(x).$$ $(\frac{dy}{dx}=2*2(tan(x))*(sec^2(x))-2*(sec(x)*tan(x)sec(x))$) $(\frac{dy}{dx}=4tan(x)sec^2(x)-2tan(x)sec^2(x)$) $(\frac{dy}{dx}=2tan(x)sec^2(x)$)