# Tutor profile: Emily M.

## Questions

### Subject: Calculus

What is the mass of a disk of radius $$4$$ and density function $$\rho(x,y)=\sqrt{x^2+y^2}$$ ?

First, let's think about how to calculate the mass. For a disk of infinitesimally small height, mass is simply density times area: $(m=\rho A$) We are given the density as a function of $$x$$ and $$y$$, so we must figure out how to calculate the mass from that density function. We must integrate over the entire area of the disk in order to calculate the mass, and given the circular nature of a disk, spherical coordinates are the best way to go. We will look at the mass equation in integral form: $(\int dm=\int \rho(x,y)dA$) Now, we need to establish was $$dA$$ is. Lets look at a tiny portion of a disk: To understand the shape of this area, draw a straight line from the inner 't' to the outer 't' in the drawing below. Draw another line from the inner '0' to the outer '0'. o o o t o o t o o o 0 0 o o o o o o o o You should have a shape made of two arc lengths, the bottom (t$$\rightarrow$$0) being smaller than the top (t$$\rightarrow$$o$$\rightarrow$$0), and two straight lines connecting those arc lengths (t $$\rightarrow$$ t and 0 $$\rightarrow$$ 0). Now, to calculate this $$dA$$, lets look at each component of the shape. The overall trend of the shape is like a rectangle, in which area is simply length times width. So lets find the length and width of our $$dA$$ shape. The width is the length from t $$\rightarrow$$ t, which equals the length from 0 $$\rightarrow$$ 0. This is a change in radius, so the width of our $$dA$$ is $(dr$) The length of our piece is the arc length connecting t$$\rightarrow$$o$$\rightarrow$$0 (We could have used the arc length from t$$\rightarrow$$0, but since this is an infinitesimally small piece, one can assume the two arc lengths to be the same) Arc length, s, is calculated by: $(s=r\theta$) So for our case, this will be an infinitesimally small piece of arc length: $(ds=rd\theta$) So now we can find $$dA$$: $(dA=dr * rd\theta$) Written more neatly: $(dA=rdrd\theta$) Lets plug this in to our integrals from earlier: $(\int dm =\int \int \rho(x,y) rdrd\theta$) Plugging in our density equation: $(\int dm = \int \int \sqrt{x^2+y^2}rdrd\theta $) We need all the variables in terms of $$r$$ and $$theta$$ since those are the variables we are integrating over. On a circle, we know that $$r^2=x^2+y^2$$ Thus, we can replace any instance of $$x^2+y^2$$ with $$r^2$$ $(\int dm = \int \int \sqrt{r^2}rdrd\theta $) which simplifies to $(\int dm = \int \int r*rdrd\theta = \int \int r^2drd\theta$) Now we must figure out the limits on our integrals: We know the radius of the disk can range from $$0$$ to $$4$$ Thus, $(\int dm = \int \int_0^4 r^2drd\theta$) Recall that in order to cover the entire disk, $$\theta$$ must range from $$0$$ to $$2\pi$$ Thus we have $(\int dm = \int_0^{2\pi} \int_0^4 r^2drd\theta$) Solving both sides gives us $(m=\int_0^{2\pi}\frac{1}{3}r^3|_0^4d\theta=\int_0^{2\pi}\frac{1}{3}(64)d\theta$) $(m=\frac{1}{3}(64)(2\pi)$) Thus, $(m=\frac{128}{3}\pi$)

### Subject: Physics

Sally-Sue is playing with a hula hoop of radius $$2$$ meters (assume negligible difference between inner and outer radius). As she plays, Sally-Sue spots a nearby ramp and decides to roll her hoop down it. This ramp has a height of $$3$$ meters, a length of $$5$$ meters, and an angle of $$\theta=35^o$$ with the horizontal. If hula hoop starts at rest from the top of the ramp and rolls without slipping, what is its velocity once it reaches the bottom?

The easiest way to solve this problem is with energies. Because we are dealing with a rotating hoop we will have to factor in moment of inertia and angular momentum - both of which make using the classical kinematic equations more complicated! As a general rule, always use energies when you can. The law of conservation of energy says that energy is neither created nor destroyed, it simply changes form. This means we can say the energy of the system at the beginning equals the energy of the system at the end. Written concisely, this is represented by: $(E_o=E_f$) So lets first look at the hoop while it is at the top of the ramp. Let's try and draw a picture of the situation: Here we have the hoop, represented by O, on top of a ramp of height $$3$$, length $$5$$, and angle $$\theta$$. O |**** $$3$$ |********** |**************** |********************$$^\theta$$** $$5$$ Let's calculate the total energy of the hoop in this state! The hoop is at rest, so we know there is no kinetic energy, $$KE=0$$. The hoop is above the ground, so it has some potential energy ($$PE$$) The formula for potential energy is: $(PE = mgh$) where $$m=$$ mass in kilograms (kg), $$g=$$ acceleration due to gravity, which is approximately $$-9.81 \frac{m}{s^2}$$ and $$h=$$ height in meters (m) For our hoop, the potential energy is $(PE = mg(3)$) because it is 3 meters above the ground. and since that is the only energy contributing to the total energy in the initial state, $(E_o=3mg$) Now, lets look at the hoop when it is at the bottom of the ramp: |**** $$3$$ |********** |**************** |********************$$^\theta$$** O $$5$$ Lets find the energy in this state: Here, the hoop has no potential energy because it is at the bottom of the ramp! Thus, $$PE = 0$$ Now, let's look at the kinetic energy: This is a rotating hoop so it has both translational kinetic energy ($$KE_t$$) and rotational kinetic energy ($$KE_r$$). So $$E_f = KE_t + KE_r$$ First, lets look at the translational kinetic energy: $(KE_t = \frac{1}{2}mv^2$) where $$m$$ is mass and $$v$$ is velocity of the hoop at the bottom of the ramp. Now, the rotational kinetic energy is: $(KE_r = \frac{1}{2} I \omega ^2$) where $$I$$ is the moment of inertia for a thin hoop and $$\omega$$ is the angular velocity $$I$$ for a thin hoop is: $(I = mr^2$) For our hoop, $$I = m2^2$$, but I will refrain from plugging in numbers until the end. Now, we must recall the relationship between angular velocity and linear velocity: $(\omega = \frac{v}{r}$) So we have $$KE_r = \frac{1}{2}(mr^2)(\frac{v}{r})^2$$, where $$r$$ is the radius of our hoop. Notice how the radius cancels out! So we end up with: $(KE_r = \frac{1}{2}mv^2$) So now we have: $$E_f = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$$ Now, putting these pieces together: $(E_o=E_f$) $(3mg=mv^2$) Now, solving for $$v$$ (notice how mass cancels out!): $(v^2=3g$) thus, $(v=\sqrt{3g}$) I want you to notice something: Neither mass nor radius of the hoop factored in, and we never had to use the angle of the ramp! This is the beauty of working with energies, it often simplifies situations into much easier problems to work.

### Subject: Algebra

Sally is standing on a beach with her dog brings her a ball. She throws the ball from the sand and its height in the air as a function of time is modeled by the following function: $(y(t)=27t-9t^2$) When will the ball hit the ground?

This is word problem, which means we are going to have to look carefully at what the question is asking and what we need to do in order to answer it. The question is asking when the ball will hit the ground. This means we are looking for the time at which the ball hits the ground. For simplicity, let's call this $$t_{ground}$$. That leads us to another question: What is the height of the ball when it hits the ground? Find something around you that is on the ground - could be a chair or your feet, anything! How high above the ground is it? Zero - because it is resting on the ground. This means the height of the ball when it hits the ground is zero. Lets make a list of things we know: $(y(t_{ground})=\ 0$) And another list of things we are looking for: $(t_{ground}=\ ?$) Since the height of the ball, $$y(t)$$, is zero when it hits the ground, let's substitute $$0$$ in for $$y(t_{ground})$$ to solve for time! $(0=27t_{ground}-9t_{ground}^2$) Now, let's solve for $$t$$. Let's first try factoring: Factoring is the process of splitting up an equation into simpler pieces. Right now, we've got this mess of $$t$$'s and $$t^2$$'s, and factoring will put this into pieces that are easier to understand. The first thing we want to notice is that each term (each piece) of our equation has a $$t$$ in it. Notice how we have a $$27t$$ and a $$9t^2$$ (which can be rewritten as $$9t\times t$$). This means we can "factor" it out! By factoring, $(0=27t_{ground}-9t_{ground}^2$) becomes $(0=(27-9t_{ground})t_{ground}$) (We can check our work by distributing the $$t_{ground}$$ back in by multiplying each term by $$t_{ground}$$). Now that we have simplified our equation to $(0=(27-9t_{ground})t_{ground}$) We can solve for $$t_{ground}$$ Something we always have to remember with an equation is that both sides of the equation are equal to the same thing. (Say I have $$4+3=x$$, $$x$$ must equal 7 because that is what the left side equals). So back to our problem, the left hand side of the equation is $$0$$. That means the right hand side of our equation must also be equal to $$0$$. Let's do some thinking... Recall that anything multiplied by zero is zero: $$5 \times 0 = 0$$ $$y \times 0 = 0 $$ That is because multiplication is like repeating: $$5 \times 2$$ is the same as having 5 two times ($$5+5$$). Thus, if you "repeat" something zero times, you have zero! Back to our equation: $(0=(27-9t_{ground})t_{ground}$) Remember that we are trying to find values for $$t_{ground}$$ such that the right hand side of the equation is zero. And we just learned that anything times zero gives us zero, so let's put those two pieces together: Our equation has two factors now: $$(27-9t_{ground})$$ and $$t_{ground}$$ They are multiplied by each other, so that means if either of those factors are equal to zero, the entire function is equal to zero: $(0\times t_{ground}=0$) and $((27-9t_{ground})\times 0=0$) So if $$(27-9t_{ground})=0$$, what is $$t_{ground}$$? $(27-9t_{ground}=0$) We want to isolate $$t_{ground}$$, So let's add $$9t_{ground}$$ to both sides which will get the $$9t_{ground}$$ on its own (Adding to both sides of the equation maintains the equality): $(27-9t_{ground}+9t_{ground}=0+9t_{ground}$) Which simplifies to $(27=9t_{ground}$) Now, we want to get rid of that $$9$$ because we want $$t_{ground}$$ all by itself. We do this by dividing by $$9$$ (Recall that $$\frac{9}{9}=1$$ and multiplying by $$1$$ does not change the value): So we divide both sides by 9 (divide both sides to maintain equality): $(\frac{27}{9}=\frac{9t_{ground}}{9}$) Which simplifies to: $(3=1t_{ground}$) Since mulitplying by $$1$$ doesn't change a value, $(3=t_{ground}$) So we just solved for one value of $$t$$ - Give yourself a pat on the back! We still have one more to solve for because our equation has two terms, but don't worry - the second one is much easier to solve for. Recalling our factored equation: $(0=(27-9t_{ground})t_{ground}$) We just solved for $$t_{ground}$$ if the first term, $$27-9t_{ground}$$, was equal to zero. Now, we need to solve for $$t_{ground}$$ if the second term is equal to zero. This becomes quite simply since our second term is only $$t_{ground}$$ Thus, $(t_{ground}=0$) So we have solved the equation for $$t_{ground}$$ and found that the values $$0$$ and $$3$$ solve the equation when the height of the ball is zero. Let's think about what these mean. At time equal to zero, what is going on? Well, that is when Sally is initially throwing the ball, so that isn't the time we are looking for. That means time equal to three is the time at which the ball hits the ground! So the ball will hit this ground at $(t=3$). Give yourself a round of applause! This was a tough problem with many steps, but its possible!

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