# gradient! urgent

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• Oct 17th 2007, 04:53 AM
seb
gradient! urgent
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
• Oct 17th 2007, 05:19 AM
topsquark
Quote:

Originally Posted by seb
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

$y = 2.4x^3 + 3.1x^2 - 2x + 1.5$

So
$\frac{dy}{dx} = 7.2x^2 + 6.2x - 2$

i) Find the gradient for x = 0 and x = 5.

ii) Find when $\frac{dy}{dx} = 0$

-Dan
• Oct 17th 2007, 05:21 AM
ticbol
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.