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Mathematics > Calculus > Flashcards

Calculus Flashcards

1
Q

A function is given by f(x) = (equation). Find the gradient of the graph of f at the point where x = (x value given).

A
  1. Differentiate the function
  2. Substitute x for (x value given) into the derived function
  3. Solve to find the gradient
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2
Q

For a function y, d(y)/d(x) = (equation). The graph of y passes through the point (2,5). Find the equation of the function y.

A
  1. Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
  2. Simplify the equation
  3. Substitute x, y values into the equation to solve for c
  4. Rewrite antidifferentiated equation with xy variables and value of c
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3
Q

(variable 1) a and (variable 2) b are given by a(b) = (equation). Find the greatest value of a.

A
  1. Differentiate the function
  2. Make function = 0
  3. Solve to find value of b
  4. Substitue value of b back into the original function
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4
Q

The graph of f(x) = (equation) has a turning point of (2,5). Find the gradient of the function at the point where x = 2.

A
  1. Differentiate the function
  2. Make function = 0
  3. Substitute value of x of the turning point into the equation to solve for the value of the other variable
  4. Replace other variable in the original equation with the value
  5. Differentiate the function
  6. Substitute value of x into the equation
  7. Solve to find the gradient
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5
Q

f(x) = (equation). For what values of x is f a decreasing function?

A
  1. Differentiate the function
  2. Make function = 0
  3. Factorise
  4. Solve for x values
  5. Solve the differentiated function for when f(0) to ‘check gradient’. If the result is < 0, it is decreasing, so write x values a domain. If the result is > 0, it is increasing, so write the x values as less than the bottom number, and more than the top number.
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6
Q

To differentiate fractions…

A
  1. Differentiate the top line of the fraction

2. Divide the result by the bottom line

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7
Q

(variable 1) a from a fixed point is modeled by the function a(b) = (equation), where b is the (variable 2) from the point. What will a be when (variable 3) c is at its minimum?

A
  1. Differentiate the equation to find the equation for c
  2. Differentiate the equation again to find the equation for d
  3. Make equation for d = 0
  4. Solve to find value of b
  5. Substitute value of b into the original equation
  6. Solve to find a
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8
Q

The graph of a function y = f(x) passes through (0,0), and its gradient function is f’(x) = (equation). Find the y-coordinate of the point on the curve where x = 2.

A
  1. Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
  2. Substitute c with y-intercept value
  3. Substitute x for 2 into the function
  4. Solve to find the y-coordinate
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9
Q

(variable 1) a, in relation to (variable 2) b can be modeled by the function a = (equation). When will it be before a will be changing at a rate of (± amount) per b.

A
  1. Differentiate the equation
  2. Make equation = (± amount)
  3. Solve to find the value of b
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10
Q

Sketching function graphs

A
  1. Parabola
    - If the parabola is curved downwards, the cubic will be negative.
    - X-intercepts of parabola are the points the cubic curves
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11
Q

The gradient of the graph of a function is given by dy/dx = (equation). At the maximum turning point of the graph of the function, y = 2. Find the equation of the graph.

A
  1. Factorise
  2. Solve for x values
  3. Use the largest x value
  4. Solve the differentiated function for when f(2) to ‘check gradient’. If the result is < 0, it is decreasing (minimum). If the result is > 0, it is increasing (maximum)
  5. Write x values as a domain
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12
Q

The gradient of the graph of a function is given by dy/dx = (equation). At the maximum turning point of the graph of the function, y = 2. Find the equation of the graph.

A
  1. Make equation = 0
  2. Factorise
  3. Solve for x values
  4. Use the largest x value
  5. Solve the differentiated function for when f(a number either side of the x value) to ‘check gradient’. If the result for the value is > 0, it is decreasing (minimum). If the result is < 0, it is increasing (maximum)
  6. State maximum coordinate as (x value from above, y value given)
  7. Anti-differentiate the equation, then include +c to the end of the equation
  8. Substitute xy values into the equation to solve for the value of c
  9. Rewrite the equation with the value of c, and the xy variables
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13
Q

The tangent to the graph of the function y = (equation) at the point where x = 2 passes through the point (2, 5). Find the value of a, and therefore the equation of the tangent.

A
  1. Find the equation when x = x value given (This is y1)
  2. Differentiate the equation
  3. Find the differentiated equation when x = x value given (This is m)
  4. x value given (This is x1)
  5. Substitute values into the tangent equation using brackets which is y - y1 = m(x - x1)
  6. Substitue xy values of point given into the equation to solve for the value of a
  7. Rewrite the tangent equation with the value of a, and the xy variables
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14
Q

Find the x coordinate of the point of the graph of the function f(x) = (equation), where the gradient is equal to 2.

A
  1. Differentiate the equation
  2. Make equation equal to 2
  3. Solve to find the value of x
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15
Q

Finding rate - with x given

(variable 1) a, in relation to (variable 2) b can be modeled by the function a(b) = (equation). Find the rate at which a is changing with respect to b, when b is 2.

A
  1. Differentiate the equation
  2. Substitute the value of b into the equation
  3. Solve to find the rate
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16
Q

Finding rate - with f given

(variable 1) a, in relation to (variable 2) b can be modeled by the function a(b) = (equation). Find the rate at which a is changing with respect to b, when a is 2.

A
  1. Make a(b) on the left side of the equation = 2
  2. Solve equation to find value of b
  3. Differentiate the equation
  4. Substitute the value of b into the equation
  5. Solve to find the rate
17
Q

The length of a cuboid is 3x its width. The sum of the height, width and length is 150cm. The volume can be expressed as V = 450x^2 - 12x^3, where width is x cm. Find the height of the cuboid for which the volume is maximum.

A
  1. Differentiate the function
  2. Make function = 0
  3. Factorise
  4. Solve for x value
18
Q

Given function and gradient function graphs, find the value m, the y-value for the maximum turning point of the function f(x).

A
  1. Find equation of gradient function graph of parabola and write as a function
  2. Expand/simplify to get function into ax^2 + bx + c form
  3. Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
  4. Show that c = 0, as the point is a turning point
  5. Remove c from the antidifferentiated function
  6. Solve the antidifferentiated function for when f(x value of turning point from function graph) to find the gradient
19
Q

Equations for parabola

A

x-intercepts -
y = ±a (x±b)(x±c)

vertex -
y = ±a (x±b)^2 ±c

20
Q

(variable 1) a, is (amount 1). This changes at a rate b of (amount). How far will it travel from when a is (amount 1), to when it is (amount 2).

A
  1. a(b) = (rate)b + c
  2. Make c = (amount 1)
  3. Antidifferentiate equation to find the equation for distance
  4. Make a(b) = (amount 2)
  5. Solve to find distance
  6. Substitue distance into the antidifferentiated equation
21
Q

(variable 1) a, is (amount 1). This changes at a rate b of (amount). How far will it travel from when a is (amount 1), to when it is (amount 2).

A
  1. a(b) = (rate)b + c
  2. Make c = (amount 1)
  3. Antidifferentiate equation to find the equation for distance
  4. Make a(b) = (amount 2)
  5. Solve to find distance
  6. Substitue distance into the antidifferentiated equation
22
Q

(variable 1) a, is (amount 1). The (variable 2) b in relation to (variable 3) c1 can be modelled by the function b(c1) = (equation). What is the greatest value of a, with respect to c1.

A
  1. As b is negative, a will have its greatest value when b = 0
  2. Make equation = 0
  3. Solve for c1
  4. Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
  5. Make c = (amount 1)
  6. Substitute the value of c1 into the antidifferentiated function
  7. Solve to find a
23
Q

(variable 1) a, is (amount 1). The (variable 2) b in relation to (variable 3) c1 can be modelled by the function b(c1) = (equation). How far does object travel before it stops?

A
  1. Stops when value of a = 0.
  2. Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation to get the equation for a in relation to c
  3. Make c = (amount 1)
  4. Make antidifferentiated equation = 0
  5. Solve to find t
  6. Antidifferentiate the function again by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation to get the equation for distance in relation to c
  7. Show that c = 0, as when c1 = 0, distance = 0.
  8. Substitute value of t into the antidifferentiated function
  9. Solve to find distance
24
Q

A function f is given by f(x) = (equation). Find the equation of the tangent at the point on the graph of f where the gradient is 0. In relation to the graph, fully describe the point where this tangent meets the function.

A
  1. Differentiate the function
  2. Make function = 0 (gradient given)
  3. Solve to find value of x
  4. Substitue value of x back into the original function to find value of y
  5. Solve the differentiated function for when f(a number either side of the x value) to ‘check gradient’. If the result for the value is > 0, it is decreasing. If the result is < 0, it is increasing.
  6. “The point (coordinate found) is the (minimum/maximum point of the curve (turning point).” We know this because the gradient of a turning point is 0, and we have checked the gradient either side of the x value
25
Q

Object is (variable 1) a at the point where (variable 2) b from a fixed point is (variable 3) c. This can be modelled by a(c) = (equation). What is the maximum value of a.

A line on the side of a curve is a tangent to it. The line can be modelled by the function y = (equation). Find the vertical distance where the tangent will meet the curve.

A
  1. Differentiate the equation
  2. At maximum height (the turning point of curve), gradient = 0
  3. Make differentiated equation = 0 (gradient)
  4. Solve for the value of c
  5. Substitute value of c back into the original equation
  6. Find gradient of straight line from in front of x value given
26
Q

A line on the side of a curve is a tangent to it. The line can be modelled by the function y = (equation). Find the vertical distance from below the top of the curve where the tangent will meet the curve.

A
  1. Find gradient of straight line from in front of x value given
  2. Make differentiated equation = gradient
  3. Solve for the value of x
  4. Substitute value of x back into the original equation
  5. Subtract the answer from the total height of the curve
27
Q

(variable 1) a at a (variable 2) b from another point c, can be modelled by the function y = (equation). Find the turning points of the curve and state whether this complies with height restrictions.

A
  1. Differentiate the equation
  2. Make equation = 0
  3. Factorise
  4. Solve for x values
  5. Substitute each x value back into the original equation
  6. Ensure that for each value of x, y does not exceed the height for the restrictions
  7. Solve the differentiated function for when f(a number either side of each of the x values) to ‘check gradient’. If the result for the value is > 0, it is decreasing. If the result is < 0, it is increasing.
  8. Determine maximum and minimum points

Mathematics (12 decks)

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