Question

The equation of a curve is y = √
(5x + 4).
(i) Calculate the gradient of the curve at the point where x = 1. [3]
(ii) A point with coordinates (x, y) moves along the curve in such a way that the rate of increase of x
has the constant value 0.03 units per second. Find the rate of increase of y at the instant when
x = 1. [2]
(iii) Find the area enclosed by the curve, the x-axis, the y-axis and the line x = 1. [5]

Answer 1

gradient mean
\[\frac{ dy }{ dx}\]

Answer 2

first calculate the dy/dx

Answer 3

\[\frac{ dy }{ dx }=\frac{ 1 }{ 2 }\left( 5x+4 \right)^{\frac{ -1 }{ 2 }}\]

Answer 4

got that far

Answer 5

loll you forgot to differentiate 5x+4

Answer 6

after differentiating the root we also differentiate what is inside it

Answer 7

okk...got it ??..this will be our gradient

Answer 8

now rate of chnage of x = dx/dt

Answer 9

differentiate y with respect with t

Answer 10

use this
\[\frac{dy}{dt} = \frac{dy}{dx}*\frac{dx}{dt}\]