Tasman B.

Jr. level Aerospace Engineering Student, Online Math Tutor for one year, and 2 years of teacher training and experience.

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Pre-Calculus

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Question:

Two ducks flew away from the same pond. The first duck flew 3 miles 40 degrees from due east, and the second duck flew 5 miles -20 degrees from due east. How far apart are the ducks from each other?

Tasman B.

Answer:

This problem involves the idea of vectors and their components. First we should look at the problem as if we were looking down on the birds flying. Then to solve we will need to break these vectors up into their respective components and then use the distance formula (or essentially the Pythagorean Theorem) to solve. In short: The first duck’s y-component: 3*sin(40) = 1.928 miles, and the first duck’s x-component: 3*cos(40)=2.298 miles. The second duck’s y-component: 5*sin(-20) = -1.710 miles, and the second duck’s x-component: 5*cos(40)=4.698. Then applying the components as if they were the x and y values for points on a graph we can simply use the distance formula to solve for how far apart they are. Thus the distance apart is (where sqrt means the square root of): sqrt{ [1.928-(-1.710)]^2+[2.298-4.698]^2 } = 3.92 miles In more detail: Imagining the ducks flight path on a graph will give us two straight lines on an angle from the positive x-axis. The components of these paths are simply the x and y values of the ducks' final position. To find these components we should draw a right triangle from the origin to our ducks’ final stopping point where the path of the bird (from origin to final position) is the hypotenuse. With these figures it is easier to see that we can use some simple trigonometry to solve for our x and y components. Trigonometric Identities Recalling that sine is the opposite side length (in reference to our angle) over the hypotenuse length we can then rearrange our formula from: sin(angle) = opposite/hypotenuse, to: hypotenuse*sin(angle) = opposite. Now we can easily solve for our y-components of our vectors (see above). Using another trig identity of cosine is the adjacent side length (in reference to our angle) over the hypotenuse length. Once we rearrange our formula from: cos(angle) = adjacent/hypotenuse, to: hypotenuse*cos(angle) = adjacent, we can solve for x-components of our vectors (see above). Distance Formula Now that we have the components of the vectors we can use the distance formula to solve how far apart the ducks are now. The distance formula, which is sqrt{ [y(1) –y(2)]^2+[x(1) – x(2)]^2 } is ultimately the Pythagorean Theorem for points on a graph. If we were to draw another right triangle from the two final positions of the ducks (with the straight distance from the two points being the hypotenuse) we can more easily see that we simply need to use a^2 + b^2 = c^2 where the a and b side lengths are either the x components or y components (depending on how you draw it and label it). Then rearranging for c we have c = sqrt(a^2 + b^2), where sqrt means the square root of. In order to find the lengths of the sides of the triangle, it is simply the difference between the point values. For example, the side length in the y-direction is: [1.928-(-1.710)], and the side length in the x-direction is: [2.298-4.698]. Thus, once we plug these into the rearranged Pythagorean Theorem we get: sqrt{ [1.928-(-1.710)]^2+[2.298-4.698]^2 } = 3.92 miles.

Trigonometry

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Question:

A rubber duck is floating on water. It bobs up and down with the waves with a motion that can be represented as a sinusoidal function. As you begin to watch (time = 0) the rubber duck is at the peak of a wave and is 10 meters above the ocean floor. The duck then sinks to the trough of the wave at pi/3 seconds later, where it is 9.5 meters above the ocean floor. Model the motion of this rubber duck in the form of D(t) = a*cos(b*t)+c where D(t) is the distance in meters the rubber duck is from the ocean floor.

Tasman B.

Answer:

Here we just need to understand 3 main trends of sinusoidal functions. They are: amplitude, midline, and period. In short: The answer to this problem is: 0.25*cos(3*t)+9.75. In Detail: Think of the amplitude ("a" for the form above), period ("b" for the form above), and midline ("c" for the form above) as modifiers to our original sinusoidal equations. Midline: The midline is where the sinsoidal function is oscillating around. In other words, it's the middle line in the path of the duck (right in the middle between the peak and the trough of the path). Normally a sinusoidal function's midline is 0, so to move it we just add the value of where we want the midline to be. In the case above, the peak is 10 meters and the trough is 9.5 meters, so right in the middle of these is 9.75. Thus for the form D(t) = a*cos(b*t)+c, c = 9.75. Amplitude: Amplitude is how high/low the sinusoidal function reaches from the midline. Or in other words its the half the total range of the rubber duck. The normal amplitude of a sinusoidal function is one, so we must multiply it by the actual amplitude we want it to be. Since the maximum height in the given situation is 10 meters, and the minimum height is 9.5 meters, that means we want the function to reach 0.25 meters above the midline and 0.25 meters below the midline. Thus for the form D(t) = a*cos(b*t)+c, a=0.25 Period: Period is how long it takes the the function to repeat itself. Since we began measuring from when the duck was at the peak of a wave, the period will be from that point until the duck reaches a peak again. Normally a sinusoidal period is 2pi, so we then have to multiply the inside of the sinusoidal function (cos(in here*t)) by a ratio of 2pi. Our ratio is simply 2pi/(whatever our actual period is). We need to be careful on deciding what our actual period is because we aren't told the time for the duck to reach the peak again, we are told the time it takes for the duck to go from peak to trough. Luckily though, that is simply half the time. Therefore our current period now is 2*(pi/3). Thus for the form D(t) = a*cos(b*t)+c, b=(2pi)/(2*(pi/3)); or simplified b = 3. *Further Info: We could actually represent this rubber duck's motion by using the form a*sin(b*(t+d))+c, but we would need to put a horizontal shift on it "d". Sine and cosine sinusoidal graphs are only different by a horizontal shift of pi/2 (moving sine pi/2 to the left would produce a cosine graph, essentially). So we may want to say then that "d" would be pi/2, but since we want it to go to the left we would say that "d" is -pi/2. Then you would place in all the same numbers for "a", "b", and "c"as we did when we solved for cosine.

Algebra

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Question:

What is end behavior of the polynomial y = ax^8 + bx^3 - c?

Tasman B.

Answer:

Finding the end behavior of polynomials is a conceptual math question rather than an actual number crunching problem. The exponents can be daunting but we really only concern ourselves with the largest exponent and its coefficient (aka the number in front of it, "a" in our case) because we are only concerning ourselves with where the function's ends (that the polynomial represents) will ultimately be heading towards: positive infinity or negative infinity. So once we understand the patterns of the exponents we can readily find the solution to any end behavior polynomial. Simply put: 1) if the largest exponent is an even number then the function (that the polynomial represents) is essentially a parabola at the extreme ends (meaning a very far distance from the origin). It will open up if the coefficient ("a" in our case) is positive, or it will open down if the coefficient is negative. So both ends will be heading towards positive infinity if the coefficient is positive or both ends will be heading towards negative infinity 2) if the largest exponent is an odd number then the function (that the polynomial represents) is essentially a line at the extreme ends (meaning a very far distance from the origin). Where the line has a positive slope if the coefficient ("a" in our case) is positive, or the line will have a negative slope if the coefficient is negative. Therefore if a is positive then the left end behavior will be heading towards negative infinity and the right end will be heading towards positive infinity; if a is negative then the left end behavior will be heading towards positive infinity and the right end will be heading towards negative infinity. Thus: In our case, since our largest exponent is 8 (which is an even number), our end behavior is that both ends will go to positive infinity if a>0 or our end behavior is that both ends will go to negative infinity if a<0 More of Why: Consider the polynomial y = x^2, this polynomial gives us a parabola that opens up, and if it was y = -x^2 we get the same parabola but this time we get negative y values instead of positive. Now as you go up in even exponents (i.e. y = x^4, y= x^6, etc.) the graphs are essentially the same but they are increasing (or decreasing if they are negative) at a quicker rate. Next consider the polynomial y = x^3, this function comes from negative infinity, crosses through the origin and then heads to positive infinity. The directions become flipped if the polynomial becomes y = -x^3. Now as you go up in odd exponents (i.e. y=X^5, y=x^7, etc.) then you get essentially the same graphs only they change at a quicker rate. Now what about when there are additional values besides the largest exponent (i.e. y=x^6+3x^3 +8x - 2), it would seem think that those values would affect how the graph would look, which is right! But as you get further away from the origin you also get bigger and bigger x values. Eventually you would get to a point where you are putting in big enough x-values that the rest of the polynomial (in the example just given, the "3x^3 + 8x - 2") has minimal effect on resulting value. This is why we call it an "end behavior." There can be lots of things going on closer to the origin, but we are only concerning ourselves with what happens at the extreme ends (when x is extremely large).

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