Hi someone please help with this one
a bicycle track can be modelled by the curve with the equation
y= 2.4x³ + 3.1x² - 2x + 1.5, 0 ≤ x ≤ 5
where x and y are respectively the horizontal and vertical distances (in km) from the origin
(i) find the gradient at the start and end of the track
(i) find when the gradient of the path is zero
THANKS
Gradient is slope, is dy/dx, is y'.
y = 2.4x³ + 3.1x² - 2x + 1.5, 0 ≤ x ≤ 5
y' = 7.2x^2 +6.2x -2
(i) find the gradient at the start and end of the track
At the start, where x=0.
m = y' = -2 -----------------answer
So the track starts going downward.
At the end, where x=6.
m = y' = 7.2(5^2) +6.2(5) -2 = 209 -----------------answer
So the track ends going upward very steeply.
(ii) find when the gradient of the path is zero
So, when y' = 0,
7.2x^2 +6.2x -2 = 0
x = {-6.2 +,-sqrt[(6.2)^2 -4(7.2)(-2)]} / (2*7.2)
x = {-6.2 +,-9.8} / 14.4
x = 0.25 or -1.1111
Only the 0.25 is in the domain, so, at x=0.25, the gradient is zero. ------answer.