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Question: Let A_n be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2 * Pi / n, show that A_n = 1/2 * n * r^{2} * Sin(2*Pi / n)
- Answers: The area of each isosceles triangle is base*h/2. The side of each triangle has a length of r, so by Cosine Law, the base can be found by: base^{2} = r^{2} + r^{2} - 2r^{2} cos(2pi/n) = 2r^{2}(1 - cos(2pi/n)). By Pythagoras theorem, h^{2} = r^{2} - (base/2)^{2} = r^{2} - base^{2}/4. Sub in the formula of base^2 from the line above, we have h^{2} = r^{2} - r^{2}/2 (1 - cos(2pi/n)) = r^{2}/2 (1 + cos(2pi/n)). As we know both the base and the height of the triangle now, we can find its area. However, to simplify the calculation a bit, I will first find the square of the area of the triangle, ie, Tri^{2} = base^{2} h^{2}/4 (so it's easier to substitute) = (2r^{2} (1 - cos(2pi/n))) * (r^{2}/2 (1 + cos(2pi/n)))/4 = r^{4}/4(sin(2pi/n)). Now take the square root, and multiply by n, to get the formula in question.
Question: A plant grows by branching, starting with one segment of length L (in the 0’th generation). Every parent branch has exactly two daughter branches. The length of each daughter branch is (2/5) times the length of the parent branch. a) Find the total length of just the 12’th generation branch segments b) Find the total length of the whole structure including the original segment and all 12 successive generations. c) Find the approximate total length of all segments in the whole structure if the plant keeps on branching forever.
- Answer: First we find the number of daughters in gen 12. Gen 1 has 2 daughters, gen 2 has 4 daughters... so gen 12 has 2^{12} daughters. Then we find the length of each daughter. If gen 0 has length L, then gen 1 has length (2L/5), gen 2 would have length (2/5)^{2} L... so each of gen 12 daughter has length L(2/5)^{12}. Total length of gen 12 is thus 2^{12} L (2/5)^{12} = L(4/5)^{12}.
For part b, we are summing (from i = 0 to 12) \sigma_{i = 0}^{12} L(4/5)^{i}. This is a geometric series, so the ans = L (1 - (4/5)^13 )/(1 - 4/5) = 5L(1 - (4/5)^{13}).
For part c, we do the same thing except we take limit to infinity. So we have \lim_{n --> \infty} (from i = 0 to n) \sigma_{i = 0}^{12} L(4/5)^{i} = 5L.
Question: Suppose I have a glass of water in a cylindrical cup of radius R. I have water filled to a height of h. I now spin the cup around the axis of the cylindrical cup. It is a known fact that the water surface assumes a paraboloid shape: it is congruent to the surface of revolution when the curve y = (a^{2} / 2g ) * x^{2} is rotated about the y-axis, where a is what is called the “angular velocity” and g is the gravitational acceleration. Find out how fast one should spin the cup (i.e. find a) in order that the water surface touches the bottom of the cup. (Assume the cup is tall enough).
- Answer: We can find the volume of water in the cup in two ways and equate them. The first way is simple: pi*R^{2}*h. The second way is to assume that the water takes the shape of the parabola in the cup, and we can use the shell method (or the disk method) to find its volume. Using the shell method (formula: int 2pi xf(x) dx), we integrate from 0 to R, int 2pi x (a^{2}/2g)*x^{2} dx = (pi*a^{2}/g) x^{4}/4 (from 0 to R) = (pi*a^{2}/g) R^{4}/4. Now we equate the two volumes to find a = sqrt(4gh/R^2).
Another way is to use the formula for surface of revolution: S = 2pi \int_a^b f(x) sqrt[1 + (f'(x))^{2}] dx.
Note that disk method would also work, but you have to be more careful. We integrate int R^{2} - x^{2} dy = int R^{2} - 2gy/asup>2 dy. The integration limit goes from 0 to the point where the parabola intersect with the side of the glass. The side of the glass is x = R and x = -R, so the intersection is y = (a^{2}/2g)R^{2}.
Question:Suppose a lake has a depth of 40 meters and is bowl-shaped, with the surface of the bowl generated by rotating the curve z = x^{2}/10 around the z-axis. Here z is height in meters from the bottom of the bowl. The distribution of sediment in the lake is stratified by height along the water column. In other words, the density of sediment (in mass per unit volume) is a function of the form s(z) = C(40−z), where z is again vertical height in meters from the point at the bottom of the lake. Find the total mass of sediment in the lake. (Your answer will have the constant C in it.)
- Answer: Basically, all we need to know is some high-school physics and know how to convert that into integration. We know mass = vol * density = area*height*density. If we slide the sediment along the z-axis (into circular disks along z-axis), then the height of this is dz. The area, which is a circle, is pi*x^{2} (x is the distance between the z-axis to the side of the bowl), and x^{2} = 10z. Now substitute this into our mass formula to have mass = pi*(10z)*dz * s(z). Put in the integral sign and integrate from 0 to 40m. We have mass = 320000Cpi/3.
Question: Alice and Bob live in the same house and take their bikes to UBC. Alice starts at 12:40 (t = 0) and accelerates for three minutes such that her speed is vA(t) = 50t (in m/min, t in min). After that she keeps a constant speed of 150 m/min. Bob got up late from his nap and starts 5 minutes later (at 12:45, i.e. t = 5). He accelerates for 3 minutes according to aB(t) = 100/3 · 2(t - 5). After that he keeps his speed constant. UBC is 2100m from their home. a) Can they both make it to the M103 lecture at 13:00? b) Does Bob catch Alice on the way? If yes, when and how far from home?
- Answer: First we find out how much Alice travel during the 20 minutes, by integrating vA(t) with respect to t. This integral has two parts, from t = 0 to t = 3, and t = 3 to t = 20. So, int_0^3 50t dt + int_3^20 150 dt = 225 + 150*17 > 2100, so she makes it. To find out the time she makes it, we solve for T such that 225 + 150*(T - 3) = 2100. T = 15.5. Next we turn to Bob, as Bob's time frame is "shifted" by 5 seconds, a way to simplify the problem is to make 12:45 "time zero" for Bob. That would make the calculations easier. In other words, from 12:45, Bob's acceleration is aB(t) = 100/3 *2t. This "time shift" is valid as long as we are not comparing Bob's traveling with others. We can find Bob's velocity in the first three minutes by integrating his acceleration: vB(t) = int 100/3 2t dt = 100/3 t^{2} + C. As we know at t = 0 (ie, 12:45), Bob's velocity is zero, so we sub this into his velocity function vB(0) = 0 to find C = 0. Knowing Bob's velocity function, we can find the distance he travels in the first 3 minutes int_0^3 100/3 (t^{2} dt = 100/3 (t^{3}/3) = 300. His velocity after the 3 minutes of acceleration is vB(3) = 300. He can arrive at school in T minutes, where T solves 300 + 300*T = 2100, or T = 6. Therefore, he arrives at school 6 minutes after he starts traveling, or 12:51.
b, Part b is simple after we use the actual time frame. Bob arrives at 12:51, and Alice arrives at 12:40 + 15.5 = 12:55:30. Bob will catch up with Alice along the way. Let the time the catch-up happens to be C minutes after 12:40 (and further assume C > 8). At that point, Alice has traveled 225 + 150(C - 3), and we equate this to the distance Bob has traveled: 300 + 300(C - 8). Solving yields C = 12.5. Thus, the catch-up happens at 12:52:30, 1650 meters away from home.
Question: You decided to put a swimming pool into your garden and for some reason you want it to have the shape of a paraboloid: Its shape is generated by rotating the function y = f( x) = ((1/4)*(x^{2}))-4 (in m) about the y-axis. You define y = 0 to be the top of the soil, therefore the pool will be 4m deep (the minimum of f(x) is at y = -4 m). Unfortunately, the former owner of your garden dumped a lot of pesticide on your lawn. The pesticide was washed into the soil and is found there with a concentration gradient c( y) = 5+y (g/m^{3}) decreasing with depth. The landfill only accepts soil with an average pesticide concentration of c0 = 5 (g/m^{3}) . After you excavated the soil for your pool, will the landfill take it or not?
- Answer: This might be a trick question. The pool goes from y = 0 to y = -4. Within this range of y, the maximum concentration of the pesticide is 5 g/m^{3} at y = 0. If the max is 5, then, the average pesticide concentration must be below 5 g/m^{3}. Thus the landfill will accept it.
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