As level MATH QUESTION. NEED GUIDANCE

[naavo.1234]

THE QUESION IS AS FOLLOWS:


1 (a) Given that y=x^3 -3x, find dy/dx. Hence obtain an expression for the approximate increase in the value of y when x increases from 2 to 2+P, where P i small.

(b) The area of rectangle has a constant value of 50cm^2. One side, of length x cm, is increasing at a rate of 0.2cm/s. Find the rate at which the other side is decreasing at the instant when x=5.

 

[soumayya]

(b) The area of rectangle has a constant value of 50cm^2. One side, of length x cm, is increasing at a rate of 0.2cm/s. Find the rate at which the other side is decreasing at the instant when x=5.

A = 50, dx/dt = 0.2, x = 5
let the other side of the rectangle be y...
Area of rectangle = x * y
50 = xy
y = 50/x
dy/dx = -50/(x^2)
When x =5, dy/dx = -2

dy/dt = dy/dx * dx/dt
dy/dt = -2 * 0.2 = -0.4

rate of decrease = 0.4

Hpe zat hlps...

 

[soumayya]

(a) Given that y=x^3 -3x, find dy/dx. Hence obtain an expression for the approximate increase in the value of y when x increases from 2 to 2+P, where P i small.

y = x^3 - 3x
dy/dx = 3x^2 - 3

When x=2, dy/dx = 9
δy/δx ≈ dy/dx
δy ≈ 9p

 

[naavo.1234]

(a) Given that y=x^3 -3x, find dy/dx. Hence obtain an expression for the approximate increase in the value of y when x increases from 2 to 2+P, where P i small.

y = x^3 - 3x
dy/dx = 3x^2 - 3

When x=2, dy/dx = 9
δy/δx ≈ dy/dx
δy ≈ 9p

what i did was i found dy/dx and substituted (2+p) instead of x i came up with a quadratic expression in the end. i was able to solve it also and found the values of p to be -1 and -3. did i do the correct method or wrong?

I did not understand the following:

δy/δx ≈ dy/dx
δy ≈ 9p

 

[hassam]

what does dy/dx = 9 means......its gradient ryt...?//// meaning that if u increase x by 1 u increased y by 9 and if u increase x by 2 u increased y by 18........similarly if u increased x by p.....u ll increase y by 9p.......i hope this simple argument clears the doubt....