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Calculus is the mathematics of all things that change. It revolves around utilizing derivatives and integrals to solve problems. To understand derivatives and integrals, consider a grid with an origin at point (0, 0). Moving one space to the left yields position (-1, 0) followed by moving two spaces up to be at space (-1, 2). Position is considered the original equation and will often show the y value as dependent on the x value. This example does not have enough information to determine y at any point as the slope has not been defined. For this example, the slope will be 3. That yields y = 3x + 5. What does that mean? The position of y begins 5 paces above the origin and will increase three paces for every increase in x. The derivative of this equation will be the rate of change, or the slope. Most equations are not this straightforward.
Consider the equation y = 3x^2 + 7. Because 7 is a constant value, it has no bearing on finding the derivative. Starting at x = 0, the first five values of y would be 0, 10, 19, 34, and 55. Because the change is not constant, a derivative must be found. To find the derivative, students would use one of two methods. The first method is the definition of a derivative.
What does this example illustrate?
This example illustrates how the constant 7 cancels itself out and h can be divided out. The result is a first order derivate in the form y’= 6x. Other ways to write this equation are y’ = y1 = dy/dx or to refer to this as velocity v or v(x). The derivative obtained can be used to determine the rate of change at any value of x. For example, when x is 0, the position is not changing at all. When x is 1, the position is changing at a rate of 6 paces upward. When x is 2, the position is changing at a rate of 12 paces upward. This method can be repeated indefinitely until the derivative is 0.
The second method is to use “rules” or “tricks” to determine the derivative. When finding the derivative, operations are maintained. That means if two values are being multiplied or added, the derivative will reflect that. The first “trick” is all lone constants have derivatives of 0. This was illustrated in the first method when the 7 cancelled itself out. The second “trick” was observed when finding the derivative of y = 3x + 5: the coefficient in front of x will remain as a constant. A parabola is more complicated because x has an exponent. That is the third “trick:” the derivative of an exponent xa will always be axa-1. In this example, the derivative of x2 will be 2x. Maintaining the operations, y’ = 3*(2x) = 0, or y’ = 6x.
What did we learn?
The first method is less memorization than the second method, but it becomes time consuming with extra work. This is undesirable during tests, especially the AP or IB exam when every second counts. The second method requires more practice to become familiar with every set of “rules” and “tricks,” but is easier to do in the long run with complicated equations.
What is integration?
Integration is the opposite of a derivative and does not have a standard definition like method one for derivation. Therefore, integration is usually learned after students are proficient in finding and utilizing derivatives. Another reason is because all the “tricks” and “rules” of integration are the opposite of derivation. Using the example y = 3×2 + 7, when a constant like 7 is integrated, it becomes a coefficient like 7x. 3×2 will be split in two parts. Remember the rule for deriving an exponent xa is it becomes axa-1 so the integral of x2 becomes x3. When deriving x3, the 3 would be dropped in front of the x. This appears in our equation 3×2 + 7 which means it does not need to be counteracted. However, if our equation was 5×2, there would need to be a coefficient of 1/3 placed before x in the form of (5/3)x3 as the derivative of that would be (5/3)*3×2. Maintaining operations and bringing the integral together, the solution takes the form:
In this example, dx is used to balance the equation and show an antiderivative due to a derivative taking the form dy/dx. dx is also used to indicate which variable was derived. This is an important distinction when taking Calculus 3 and working with equations that have more than one variable such as 3D coordinate systems.
Integrals also act as the summation of all values in a certain range or the area under the curve. A basic application of this is going pouring paint along a diagonal line down a wall and wanting to know the area of the wall that will be covered in paint. The diagonal line is in the shape y = 7x starting at the four inch mark and ending at the 9 inch mark. The integral of x1 is (1/2)x2. Multiplying by the coefficient 7 yields:
This equation must then be evaluated for the range 9 to 4. This can be done by simply plugging 9 and 4 into the equation and finding the difference of 227.5 square inches. This can be confirmed by recognizing the diagonal line forms a trapezoid with the floor and using the equation for area of a trapezoid, h(a+b)/2. Because the trapezoid is not even, the triangle cut off by the floor must be subtracted. The horizontal distance h covered by the diagonal line is 5. At marker 4, the line is 28 inches off the ground and at marker 9, the line is 63 inches off the ground. Solving, the area is 227.5 square inches. In this scenario, simply finding the integral of the line was faster and easier than finding the values of a, b, and h.
As a student advances through calculus, they’ll learn about derivatives, integrals, the practical applications, convergence theories, advanced integration techniques such as integration by parts, vector fields, and coordinate systems such as polar and spherical, multivariable calculus. Trigonometry is used heavily throughout traditional calculus. Calculus is a prerequisite for every engineering and medical student. Medical students can stop after Calculus 1 while engineering students will be expected to complete the series plus differential equations. Many finance and business programs require calculus but offer a special course that does not include trigonometry.
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The nature of calculus makes memorizing solutions impossible and demands that students do more practice problems more than they normally would.
At Miles Smart Tutoring, we are equipped with highly trained and educated individuals who can reinforce vital topics, provide guidance while completing practice problems, and teach problem-solving techniques to help students optimize time during exams.
Our tutors can help students practice for the AP Calculus Exams. All students in AP Calculus classes will take the exam at the end of their school year in hopes of earning college credit. Students in other calculus tutoring courses are not automatically enrolled in the exam but can still sign up to take it. Passing the AP exam will save students hundreds, if not thousands, of dollars in tuition, fees, and textbooks in addition to freeing up four credit hours of space for other classes.
Our calculus tutors have knowledge that extends past doing well in BC Calculus and can also provide insight to career paths that require calculus. The high caliber of our tutors means each student will obtain instruction without hesitation and with as much detail as the student’s needs demand.We also offer group tutoring sessions for calculus.
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