Starting from the base of a hill, a car drives (in a straight line) to the end of a road located at its summit. If the starting position of the car is defined by the coordinate <70, 10> and the hill peak (where the road ends) has coordinates <95, 60>, what is the gradient of the hill?
The gradience is defined by the formula:
So this formula is really just:
And given the 2 vectors:
then we simply take the y values which are:
and the x values which are:
then the formula is:
So the gradient is 2.
You do not need to memorise this formula. It is given to you in the exam. Do not memorise this formula.
When the vehicle has climbed up to a height of 50 metres on the hill (i.e. y = 50) what is the horizontal distance that has been travelled at this point (i.e. the corresponding value of x)?
- 82 metres
- 84 metres
- 86 metres
- 88 metres
- 90 metres
So in the first question we found out that they started at <70, 10> and then they vertically travel 50 metres. The final destination is <95, 60>. The current coordinate when they travel 50 metres up is <x, 50>.
The gradient stays the same.
The 60 and 50 are the:
the 95 and x are the :
But since we know the gradience is 2 and we are using the gradience formula we know that this equation must be equal to 2 (since the gradience does not change):
And then some simple algebraic manipulation gets us:
So the coordinate when it is 50 metres up is:
The following question refers to the setting described below.
An essay marking scheme is revised so that + grades can be given. Now the ordered range of grades is:
These are converted to basic marks ranging from 0 (for G) through to 6 (for A) but with + grades scored 0.5 more than the grade which the + qualfiies for (eg F+ is 1.5, C+ is 4.5 and so on).
At the same time a new formula to convert basic marks (in the range 0 to 6) to final marks is adopted. This is the formula in which:
What grade does an essay need to be awared to achieve the maximum possible final mark?
The terms maximum and minimum come from turning points in calculus.
A turning point is the point at which the line becomes horizontal. Look at this gif I made:
The line becomes horizontal at some point as it turns around the graph.
The gradient of the line touching the point <x, f(x)> is 0.
The gradient of the line touching the point <x, f(x)> is described by the function f’(x): the first derivative of f(x).
So the turning points of f(x) are those values, x, for which f’(x) = 0.
You calculate the first derivative of this function.
To check whether a turning point, t, is a local minimum or maximum you need to find the second derivative of f(x), f’‘(x) by constructing the first derivative of f’(x).
Find the value of f’‘(t) then if f’‘(t) is less than 0 then t is a local maximum. If f’‘(t) > 0 then t is a local minimum.
We get given this in the exam:
Let’s calcualte a diffrent polynomial first:
We need to find the first derivative.
The first thing we’re going to do is subtract 1 from the cube index, add that to the coefficient 2 (now 3) and times that by 2
Now we get 2 * -18 which is -36x
Now for 30x. The first derivative of 30x is just the value 30.
then we apply the polynomial rule again:
2x^3 - 18x^2 + 30x + 40
Consider the three matrices, P, Q, and R below:
- P only
- Q only
- R only
- P and Q only
- All three matrices are singular
So anytime you see a row / column of 0’s in a matrix it’s singular unless the 0’s look like this: