# Tutor profile: Denis K.

## Questions

### Subject: Physics (Newtonian Mechanics)

A projectile of 5kg is thrown at a velocity of 25m/s, 30 degrees relative to the ground, at a height of 5m above the ground. On this day, there is a constant 10N wind resistance that blows against the projectile at an angle of 15 degrees parallel to the ground. How much time does the projectile spend in the air, and how far does it travel?

In order to answer this question, we can split everything into horizontal and vertical components. But first, we must find the "deceleration" of the projectile caused by the wind. Since we have the mass of the object, the acceleration of the wind is simply: $(a_{wind}=\frac{F_{wind}}{m_{projectile}}=\frac{10N}{5kg}=2\frac{m}{s^2}$) Now, we can find the horizontal and vertical components of this acceleration: $(a_{wind,x}=-a_{wind}\cos(15^o)=-2\cos(15^o)=-1.93\frac{m}{s^2}$) And $(a_{wind,y}=a_{wind}\sin(15^o)=2\sin(15^o)=0.52\frac{m}{s^2}$) Let's proceed by finding the total time the projectile spends airborne. First, notice that the resultant vertical acceleration of the projectile is now: $(a_{res}=g+a_{wind,y}=-9.8+0.52=-9.28\frac{m}{s^2}$) Next, we can use the following equation of motion to find the total time: $(\Delta y=v_{initial,y}t+\frac{1}{2}a_{res}t^2$) We know that: $(v_{initial,y}=v_{initial}\sin(30^o)=25\sin(30^o)=12.5\frac{m}{s}$) And the change in height is $$\Delta y=0 - 5$$, therefore: $(-5 = 12.5t -4.64t^2$) So we get a time of $$t=3.05s$$. Next, we have to find the total horizontal displacement. In this case, the projectile will DECELERATE, so the appropriate equation we can use to find total displacement is: $(\Delta x = v_{initial,x}t + \frac{1}{2}a_{wind,x}t^2$) Where; $(v_{initial,x} = 25\cos(30^o)=21.65\frac{m}{s}$) And, of course $$t=3.05s$$. Therefore, we get: $(\Delta x= 21.65\cdot3.05+\frac{1}{2}\cdot-1.93\cdot3.05^2=57.06m$)

### Subject: Calculus

Compute the derivative of the following function, using: 1) The Fundamental Theorem of Calculus. 2) Directly integrating the function. $(f(x) = \int_{x^3}^{3} \frac{lnt}{t} dt$)

Before using the fundamental theorem of calculus, we must manipulate the integral using certain properties. First, we can rewrite: $(f(x) = \int_{x^3}^{3} \frac{lnt}{t}dt = -\int_{3}^{x^3} \frac{lnt}{t}dt$) We cannot simply apply the FTC in this case, because we have $$x^3$$ as the upper bound. So let us suppose another function, $$F(x)$$, which we will define to be the ANTI-DERIVATIVE of $$\frac{lnx}{x}$$. Therefore, we can apply the FTC to evaluate the integral as follows: $(-\int_{3}^{x^3} \frac{lnt}{t}dt=-\Big[F(x^3)-F(3)\Big]$) Once we have this, we can simply differentiate both sides of the equation: $(\frac{d}{dx}\bigg(-\int_{3}^{x^3} \frac{lnt}{t}dt\bigg)=-\bigg[\frac{dF(x^3)}{dx}-\frac{dF(3)}{dx}\bigg]$) Now since $$F(3)$$ is a constant, then $$\frac{dF(3)}{dx}=0$$. Furthermore, since by definition $$F(x)$$ is the anti-derivative of $$\frac{lnx}{x}$$, this mean that $$\frac{dF(x)}{dx}=\frac{lnx}{x}$$. But we have $$F(x^3)$$, so we need to use the chain rule to differentiate. This implies: $(\frac{d}{dx}\bigg(-\int_{3}^{x^3} \frac{lnt}{t}dt\bigg)=-\frac{ln(x^3)}{x^3}\cdot\Big(x^3\Big)'=-\frac{3ln(x^3)}{x}$) So our final answer is: $(f'(x) = -\frac{3ln(x^3)}{x}$) For part 2), we have to evaluate the integral first. So: $(f(x) = \int_{x^3}^{3} \frac{lnt}{t}dt=\frac{(lnx)^2}{2}\Bigg|_{x^3}^{3}=\frac{(ln(3))^2}{2}-\frac{(ln(x^3))^2}{2}=-\frac{ln^2(x^3)}{2}+\frac{ln^2(3)}{2}$) Finally, we just need to differentiate $$f(x)$$ as follows: $(f'(x) = -\frac{d}{dx}\frac{ln^2(x^3)}{2}+\frac{d}{dx}\frac{ln^2(3)}{2}=-\frac{2ln(x^3)}{2}\cdot\frac{1}{x^3}\cdot3x^2=-\frac{3ln(x^3)}{x}$) Clear,y we obtain the same answer, which is expected!

### Subject: Algebra

Paul and Mary have decided to play a little game. They will ask their good friend Michael to put a random amount of marbles in each of their bags, without letting them know how many marbles each of them will have. In order for them to figure out how many marbles are in each bag, they must collaborate! Michael has given them the following clues: 1) Upon the removal of 6 marbles from Paul's bag, his bag will have twice as many marbles as Mary's bag IF she added 3 marbles. 2) The total amount of marbles is equal to three times the number of letters in Paul and Mary's names combined. With these clues, it is possible to determine how many marbles are in each bag. Can you help Paul and Mary?

In order to answer a question like this, it is important to identify and define each variable. In this problem, we can define the following two variables: $$x:$$ amount of marbles in Paul's bag $$y:$$ amount of marbles in Mary's bag Now we can work with the given clues to create algebraic equations involving these two variables. From clue 1), we are told that upon removal of 6 marbles from Paul's bag, that is: $$x-6$$ Paul's bag will contain twice as many marbles as Mary's bag IF she added 3 marbles. So we add 3 marbles to Mary's bag: $$y+3$$ And we know at this point that Paul's bag will have twice as many marbles as Mary's bag, therefore: $$x-6 = 2(y+3)$$ Which simplifies to the following equation: $$x-2y=12$$ From clue 2), we know that the total number of marbles, that is: $$x+y$$ Is equal to THREE times the number of letters in both of their names combined! Paul has 4 letters in his name, so does Mary. This gives a total of 8 letters. This means: $$x+y=3\times8$$ Now we have two equations to solve: $$(1):x-2y=12$$ $$(2):x+y=24$$ We can solve equation (2) for y: $$y=24-x$$ And substitute into equation (1): $$x-2[24-x]=12$$ $$x-48+2x=12$$ $$3x=60$$ $$x=20$$ And to find y we can use equation (2): $$y = 24-x = 24-20 = 4$$ This gives us $$x=20$$ $$y=4$$ Therefore, Paul has 20 marbles in his bag, and Mary has 4 marbles in her bag!

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