Integrals can be applied to real life situations to solve problems relating to acceleration, velocity, and position. s(t) represents the position of a particle, s'(t)=v(t) is the velocity of the particle and s''(t)=v'(t)=a(t) is the acceleration of the particle. The integral of a(t) is equal to v(t)+c and the integral of v(t) is equal to s(t)+c. Velocity can be interpreted by a graph of the function. Where the function is positive represents the particle moving right, while where the function is negative represents the particle going left. When x=0, the particle is stopped. When asked to find the total distance traveled in a problem, you must take the absolute value of the integral. When a problem asks for the net distance or the displacement, this isn't necessary.
For example, the problem v(t)=t^2-8/(t+1)^2 asks to find the total distance traveled by the particle from 0 to 5 seconds. This can be found by calculating the integral of the absolute value of t^2-8/(t+1)^2 dt from 0 to 5, which is 44.
0 Comments
This week we learned how to make a slope field. A slope field is a graphical way of interpreting a differential equation. For example for dy/dx = x, the slope field can be graphed by plugging in values for x to find the slope at that point. This is what the slope field would look like. To find the slope field of an implicitly defined function it is easier to first write out a table, and then graph the slopes: We also learned how to solve separable differential equations this week. Dy/dx=(xy)^2 is an example of this type of problem. To solve this problem, you first need to get y and dy one the same side and x and dx on the other. Then, you can anti-derive both sides and then plug in the given point (1,1) to solve for c. The final answer to this problem would be y^(-1)/-1=x^3/3-4/3.
This week we learned how to antiderive with u substitution. Antidifferentiation with u substitution is where you select a part of the equation to be "u" and plug in u for that part. You then find du so that parts of the integral will drop out, along with dx in the original. Both definite and indefinite integrals can be computed using this method. An indefinite integral is a family of functions whereas a definite integral produces a specific number. This is an example of an indefinite integral that results in a family of equations. In this example, cos x was determined to be the u. Du is then found by taking the derivative of u. In this case, du is -sin x dx. The negative is then divided over to the other side so the cos x can be dropped from the original equation. The integral can then be written as the negative integral of e^u du. We can now solve the integral to get -(e^u + c). Finally, we substitute the u for cos x to get the final answer, -e^(cos x) + c. This is an example of a definite integral, which can be solved to find a number. This one is solved similarly to the indefinite integral. We picked (x-4) to be u and found du to be dx. We then wrote the integral in terms of u which was u^2 du. The difference between finding an indefinite integral and a definite is the bounds. When finding a definite integral with u substitution, we also have to find the new bounds. These can be found by plugging the original bounds into the equation for u. The new lower bound is -4 and the new upper bound is 0. The equation can then be solved the same way it was with the indefinite integral, but can then be evaluated from the lower bound to the upper bound to get the answer of 64/3. I think I learned the fundamental theorem of calculus primarily with inductive reasoning. I think I already knew of the fundamental theorem without actually knowing it, based on what I had already learned about derivatives and anti-derivatives. I already had the realization that an anti-derivative was the inverse of a derivative and a derivative was the inverse of an anti-derivative. The integral of a function is the same as taking the anti-derivative of a function, so an integral is also the inverse to the derivative. The word fundamental suggests that this theorem is the base for calculus, and is the primary rule in calculus. the theorem links the concept of the derivative of a function with the concept of the function's integral. I think the fundamental theorem is fundamental because everything I've learned in calculus so far has been about derivatives and anti-derivatives, which has all come from the theorem. The notation of the integral is the same as the notation used when taking an anti-derivative, which I have already learned about, which made learning about integrals easier.
This week we learned about integrals. We started by learning how to write the definite integral of a continuous function. For example, the limit as h approaches infinity of the Riemann sum of f(ck)*delta x can be written as the integral of f(x) from a to b. In this notation, a is the lower bound of the function, while b is the upper bound. The "S" is called the integral sign, f(x) is the integrand, and dx is the variable of integration. We then learned how to evaluate integrals by calculating the net area under the curve. The net area is equivalent to the area above the x-axis, which is positive, plus the area below the x-axis, which is negative. The area can be computed with a calculated by using FnInt, and typing in the integral. Integrals can be evaluated without a calculator by using the rules for definite integrals and anti-derivatives: The definite integral can also be found using anti-derivatives. The integral of f(x) from a to b is equal to f(b) minus f(a). In the first example, the integral of sin(x) from 0 to pi is anti-derived to -cos(x) and is evaluated from 0 to pi. The integral can then be computed by subtracting f(a) from f(b). So, -cos(pi) - (-cos(0)), which equals 2.
This week, we learned about related rates. Related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. The rate of change is usually written with respect to time. The process of finding a related rate can be done by following these steps:
In this problem, the goal is to find the rate at which the radius of the balloon of the balloon increases. To start the problem, a general model of a sphere is drawn, Then a mathematical model is found that corresponds to the model. In this problem, the formula for the volume of a sphere, V=4/3pi*r^3, is used as a model. The derivative of this model is then taken, which is dv/dt=4pi*r^2*dr/dt. Then everything we know that was stated in the problem is written down. These values are then plugged into the derivative and we solve for dr/dt to find that the rate at which the radius increases is .0035 cm/min.
This week was spent reviewing and relearning topics covered in pre-calculus last year. I struggled a little with remembering information learned in the second trimester of last year after not having done math since then, as I did not take the third trimester of the class. A review packet covering a large amount of topics from pre-calc was assigned as review, with a quiz over it scheduled for Tuesday. As I worked on the packet I realized I had forgotten a big portion of the things I had learned in class last year, like natural logarithms and decomposing composite functions. However, being able to work together with a group of other students on the packet, we could share what we knew and help each other complete it. The above problems are examples of decomposing composite function problems. Problem A was solved by thinking of two equations that could be plugged in to f(g(x)) to equal the root of (x^(4)-3x+1). f(x) was found to be the root of x and g(x) was found to be x^(4)-3x+1.
|
|