# Tutor profile: David P.

## Questions

### Subject: Calculus

Find the slope of the tangent line at the point $$x=\dfrac{\pi}{2}$$ of the function $$f(x) = x^2 + \cos(x)$$.

Finding the slope of a tangent line of a function evaluated at a point, simply requires us to find the derivative of the function, $$f'(x)$$, which describes the slope, or rate of change, of a function at any point $$x$$. So since we are given a specific point, we need to first find $$f'(x)$$ and then plug in the point, and this will return the slope of the tangent line at this point. In order to find the derivative: $$f'(x) = \dfrac{df}{dx} = \dfrac{d}{dx}\left( x^2 + \cos(x)\right)$$ Because the two terms of the function are added together, we can take each terms derivative separately. $$\dfrac{d}{dx}\left( x^2 + \cos{x}\right) = \dfrac{d}{dx}\left(x^2\right) + \dfrac{d}{dx}\left(\cos{x}\right)$$ $$\dfrac{d}{dx}\left(x^2\right) = 2x$$ $$\dfrac{d}{dx}\left(\cos(x)\right) = -\sin{x}$$ $$f'(x) = \dfrac{d}{dx}\left( x^2 + \cos{x}\right) = 2x - \sin{x}$$ Now that we have $$f'(x)$$ we simply need to plug in the point given into the derivative to find the slope. $$f'\left(\dfrac{\pi}{2}\right) = \pi - \sin{\dfrac{\pi}{2}} = \pi - 1$$ Therefore, $$\pi - 1$$ is the slope of the tangent line at the point. If you wanted to find the slope of the tangent line at a different point you could simply plug in the new point into $$f'(x)$$.

### Subject: Physics

A driver is approaching a red light while texting. They are currently travelling at $$25 m/s$$, and are $$100 m$$ from the red light, when they look up and notice they have to stop. Once the driver slams on the brakes, the car decelerates at $$3 m/s^2$$. Do they stop in time? How much time (in seconds) does it take for the car to come to a complete stop?

In order to determine whether the driver stops in time, we need to know how long it takes for the car to decelerate, and how far the car travels in this time. If the total distance travelled by the vehicle before coming to a stop is greater than $$100 m$$ then the car will not be able to stop in time. In order to solve this problem, let's take a look at the kinematic formulas. $$v = v_0 + at$$ $$\Delta x = \dfrac{v + v_0}{2}t$$ $$\Delta x = v_0t + \dfrac{1}{2}at^2$$ $$v^2 = v_0^2 + 2a\Delta x$$ We'd like to find a kinematic equation which relates the velocity of the car, the acceleration and the distance. Looking at the 4th equation, we can see that this would be the simplest way to solve the problem. The 4th equation uses 4 variables, $$v_0$$, which is the initial velocity, $$v$$, which is the final velocity, $$a$$, which is the acceleration, and $$\Delta x$$ which is the distance travelled. We know the final and initial velocities of the vehicle, as well as the acceleration so we can solve for $$\Delta x$$ which will tell us whether or not the car stops in time. $$0^2 = 25^2 + 2(-3)\Delta x$$ $$\dfrac{-625}{-6} = \Delta x$$ $$\Delta x = 104.1\bar{6} m$$ The distance travelled by the car is greater than $$100m$$ which means the car doesn't stop in time. Let's find out exactly how long it took the car to stop moving. For this, we just need to solve one of the kinematic formulas for $$t$$. The first one is probably the easiest, so let's use that one. $$0 = 25 + (-3)t$$ $$t = \dfrac{25}{3}$$ $$t = 8.\bar{3} s$$ At the rate at which the car decelerates when the driver hits the breaks, it takes around 8 seconds for the car to come to a stop, and the car travels further than $$100 m$$ during this time, which means the car actually runs the red light. Don't text and drive!

### Subject: Algebra

Solve the following equation for $$x$$: $$2x = x^2 - 3$$

In order to solve for $$x$$ in the equation: $$2x = x^2 - 3$$, we can actually use two different methods. Let's start with the simpler method first. When dealing with quadratic equations, we typically rearrange the terms to match the form $$Ax^2 + Bx + C = 0$$, so let's rearrange our problem to match this. By subtracting $$2x$$ from both sides of the equation we have, $$0 = x^2 - 2x - 3$$. Since this is just an equality expression, we can of course swap the two sides of the equation, $$x^2 - 2x - 3 = 0$$. Now that we have our equation in the correct form, we can either solve for $$x$$ by factoring, or by using the quadratic equation. Let us first check to see whether the equation is factorable. In order to do so, we need to determine whether there are two numbers which multiply to give $$-3$$ and add to give $$-2$$. These are the coefficients $$C$$ and $$B$$, respectively. As it turns out, $$-3$$ and $$1$$ multiply to give $$-3$$ and add to give $$-2$$, so you can factor this formula using these two values as follows, $$(x-3)(x+1) = 0$$. Notice how if you were to multiply this together using the FOIL method, you'd regain the original form above. This means that the equation has two solutions, $$x=3$$ and $$x=-1$$, as both of these values would satisfy the equation. Try it for yourself, by entering the numbers into the equation one at a time, to see if you get $$0=0$$. Sometimes, factoring an equation can be very challenging, and when this is the case you can always use the quadratic formula. As a reminder, the quadratic formula looks like this, $$x=\dfrac{-B \pm \sqrt{B^2-4AC}}{2A}$$ Let's plug in our coefficients, and verify that we still get the solutions $$3$$ and $$-1$$. $$x = \dfrac{2 \pm \sqrt{4 - 4\cdot1\cdot(-3)}}{2\cdot1}$$ $$x = \dfrac{2 \pm \sqrt{16}}{2}$$ $$x = \dfrac{2 \pm 4}{2}$$ $$x = 1 \pm 2$$ $$x = 3, -1$$ Notice how with the quadratic formula, the results are the same as with factoring. Remember that the quadratic formula will always work, but checking for easy factors is usually faster, and having multiple ways to solve a problem means you have a better chance of finding the solution, or at least checking your work.