Differentiation Questions Part 2 Flashcards
How do you find approximate value for a gradient of a curved graph?
Find P2, fairly close to gradient we’re trying to find like (1, 1), with a straight line passing through both points as an approximation for a tangent to the graph at both P1/P2, whilst the second point an approximation for the tangent to the graph at (1,1), calculating that tangent (rise/run) approximates gradient of graph at P1.
How do you ensure best approximation of curved graph gradient for a P1?
By choosing P2 as close to P1
What is a mathematical example of calculating curved graph gradient?
Calculating approximation to gradient at (1, 1), given equation on graph (x^2), the y co-ordinate would be 1.2^2 or 1.21; so p1 is (1.1, 1.21) where using rise/run gives 2.1(the approximate value for the gradient of the graph AT (1 , 1)
What is H used as in differentiation?
It is used as a variable denoting increase in x-coordinate from P1 to P2, being either positive or negative, but never 0
Why can’t H be 0?
Because P2 can be both right or left of P1 but never P1 itself.
What would X and Y of P2 be in relation to P1 being (1 , 1) and a graph y = x^2 and how would this look under RISE/RUN?
(1 + h, (1 + h)^2) OR (1 + h)^2) - 1 / (1 + h) - 1, which is simplified to 2 + h
What happens to value of H as P2 gets closer to P1?
It gets, relatively, closer to zero.
How can you mathematically express limit?
Limit of 2 + h as h tends to zero is 2
How can you apply method of finding gradient of a given P1 to finding gradient of y = x^2 at a general point, where X is denoted by X?
If P1 is (x, x^2) then P2 would be (x + h, (x + h) ^2), in which RISE/RUN equates 2x + H(when simplified