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Mathematics > Calculus (Simplified :)) > Flashcards

Calculus (Simplified :)) Flashcards

1
Q
  1. Given equation and a value for x, find the gradient
A
  1. Differentiate f(x) to get f’(x)

2. Substitue in x value into f’(x) to find the gradient

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2
Q
  1. Given equation and a value for y, find the gradient
A
  1. Substitute y value into f(x) to find x
  2. Differentiate f(x) to get f’(x)
  3. Substitute in x value into f’(x) to find the gradient
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3
Q
  1. Given equation and coordinates of a point, or only the a value, find the equation of a tangent
A
  1. Substitue x value into f(x) to find y
  2. Differentiate f(x) to get f’(x)
  3. Substitute a value into f’(x) to find the gradient of the tangent m
  4. Substitue into y - y1 = m(x - x1)
  5. Simplify into y = mx + c
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4
Q
  1. Given equation and gradient, find the coordinates of a point
A
  1. Differentiate f(x) to get f’(x)
  2. Make f’(x) = gradient
  3. Solve to find x value
  4. Substitue x value into f(x) to find y
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5
Q
  1. Given equation, find turning points and determine nature
A
  1. Differentiate f(x) to get f’(x)
  2. Make f’(x) = 0
  3. Solve to find x value(s)
  4. Substitue x value(s) into f(x) to find y
  5. Solve the f’(x) 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.
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6
Q
  1. Given equation, find where functions are increasing or decreasing
A
  1. Differentiate f(x) to get f’(x)
  2. Make f’(x) = 0
  3. Solve to find x values
  4. Solve the f’(x) for when f(a number either side 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.
  5. Write as a domain with middle value determining whether it is increasing or decreasing where (lower x value) < x < (higher x value)
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7
Q
  1. Rates of change - Given the value of x
A
  1. Differentiate f(x) to get f’(x)
  2. Substitue x value into f’(x) to find y
  3. Write as, when x is (__), y is (increasing/decreasing) at (__).
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8
Q
  1. Rates of change - Given the value of y
A
  1. Substitute y value into f(x) to find x
  2. Differentiate f(x) to get f’(x)
  3. Substitue x value into f’(x) to find y
  4. Write as, when y is (__) and x is (__), the y is
    (increasing/decreasing) at (__).
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9
Q
  1. Rates of change - Given the value of dy/dx (rate)
A
  1. Differentiate f(x) to get f’(x)
  2. Make f’(x) = rate
  3. Solve to find x value
  4. Write as, y is (increasing/decreasing) at (rate) when x is (__).
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10
Q
  1. Optimisation
A

Note - At maximum and minimum points, gradient = 0

  1. Differentiate f(x) to get f’(x)
  2. Make f’(x) = 0
  3. Solve to find x value
  4. Substitue x value into f(x) to find y
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11
Q
  1. Optimisation with related variables
A
  1. Draw a diagram
  2. Identify what needs to be optimized and write an equation
  3. Write an equation which relates x and y
  4. Substitute for y
  5. Differentiate f(x) to get f’(x)
  6. Make f’(x) = 0
  7. Solve to find x value
  8. Substitue x value into f(x) to find y
  9. Test solution then answer question
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12
Q
  1. Antidifferentiation
A
  1. Add 1 to the power
  2. Divide the constant by the new power
  3. Add +c to the end of the equation
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13
Q
  1. Antidifferentiation - Given a point on the original curve, calculate c
A
  1. Antidifferentiate f’(x) to get f(x)
  2. Substitue x and y values into f(x) to find c
  3. Rewrite f(x) with xy variables and value of c
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14
Q
  1. Antidifferentiation - Given a pair of values in a rate of change problem, calculate c
A

Note - The pair of values will be different variables (eg. How deep was the water after 10 minutes? - The variables are depth and time)

  1. Antidifferentiate f’(x) to get f(x)
  2. Substitue initial ‘x’ and ‘y’ values into f(x) to find c
  3. Substitue final x or y value into f(x)
  4. Solve to find the other x or y value
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15
Q
  1. Antidifferentiation - Given a graph of the gradient function and point it passes through, sketch the original curve
A
  1. Write the equation for the line
  2. Antidifferentiate f’(x) to get f(x)
  3. Substitute coordinate of point into f(x) to find c
  4. Calculate the minimum/maximum value by solving f(x) for when x = the x-axis intercept
  5. Draw parabola with the turning point in line with x-axis intercept and curving upwards if gradient of line is positive, or curving downwards if gradient of line is negative
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16
Q
  1. Kinematics
A
  • To convert from distance to velocity, differentiate
  • To convert from velocity to acceleration, differentiate
  • To convert from acceleration to velocity, antidifferentiate
  • To convert from velocity to distance, antidifferentiate
17
Q
  1. Differentiation in kinematics - How far above ground when object is launched?
A

Note - When something is launched, t = 0

  1. Substitue x value (t = 0) into f(x) to find y
18
Q
  1. Differentiation in kinematics - How far did the object travel during the first second of flight?
A
  1. Substitue x value (t = 1) into f(x) to find y
  2. This will give the total distance. Subtract answer for how far above ground the object was when launched from the total distance to find distance travelled in 1 second
19
Q
  1. Differentiation in kinematics - How fast was the object traveling after 2 seconds?
A
  1. Differentiate f(x) to get f’(x)

2. Substitue x value (t = 2) into f’(x) to find y

20
Q
  1. Differentiation in kinematics - What is the objects acceleration after 2 seconds?
A
  1. Differentiate f’(x) to get f’‘(x)
21
Q
  1. Differentiation in kinematics - What is the maximum height reached by the object?
A

Note - When something is at maximum height, v = 0

  1. Make f’(x) = 0
  2. Solve to find x value
  3. Substitue x value into f(x) to find y
22
Q
  1. Differentiation in kinematics - When was the velocity of the object equal to 12m/s?
A
  1. Make f’(x) = 12

2. Solve to find x value

23
Q
  1. Anti-differentiation in kinematics - Antidifferntiating once to find c - Given the accleration/velocity, calculate the velocity/distance
A
  1. Antidifferentiate f’(x) to get f(x)
  2. Substitue initial ‘x’ and ‘y’ values into f(x) to find c
  3. Rewrite f(x) with xy variables and value of c
  4. Substitue final x or y value into f(x)
  5. Solve to find the other x or y value
24
Q
  1. Anti-differentiation in kinematics - Antidifferntiating twice to find c and c’ - Given the accleration, calculate the distance
A
  1. Antidifferentiate f’‘(x) to get f’(x)
  2. Substitue initial ‘x’ and ‘y’ values into f(x) to find c
  3. Rewrite f(x) with xy variables and value of c
  4. Antidifferentiate f’(x) to get f(x)
  5. Substitue initial ‘x’ and ‘y’ values into f(x) to find c
  6. Rewrite f(x) with xy variables and value of c
  7. Substitue final x or y value into f(x)
  8. Solve to find the other x or y value

Mathematics (12 decks)

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