Equation of a curve = `y=(x-1)(x^2-6x+2)` given that z=y^2 and that z is increasing at the constant rate of 10 units per second, find the rate of change of y when x=2 ?  How do i find a rate of...

Equation of a curve = `y=(x-1)(x^2-6x+2)`

given that z=y^2 and that z is increasing at the constant rate of 10 units per second, find the rate of change of y when x=2 ? 

How do i find a rate of Change of a curve? 

Asked on by kimmiek

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gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

Given that rate of change of z = 10 unit/sec

i.e., dz/dt = 10

Also, `z=y^2`

which means,  dz/dy = 2y

multiplying the numerator and denominator by dt, we get

`(dz/dt)/(dy/dt) = 2y`  

or, `dy/dt = 1/(2y) dz/dt = 1/(2y) * 10 = 5/y`

we have to determine the rate of change of y (dy/dt) when x=2

at x=2, y = (x-1)(x^2-6x+2) = (2-1)(2^2-6*2+2) = -6

which means, dy/dt = 5/(-6) = -5/6

The rate of change of curve can found out by using derivative, in this case the first derivative of y. In general, the term w.r.t to which the rate of change has to be calculated is given in the question. For example, the rate of change of equation with respect to x can be calculated as dy/dx. Rate of change with respect to time can be calculated as dy/dt (for a given time), etc.

Hope this helps. 

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

Given dz/dt = 10 a constant

To find dy/dt at x= 2:

Z = y^2, given.

Therefore, dz/dt = 2y* dy/dt.

Therefore, dy/dt = (dz/dt)/ (2Y) = 10/(2Y) = 5/y

dy/dt = 5/y..............(1)

y = (x-1)(x^2-6x+2), given.

At x = 2, y = (2-1)(2^2- 6*2+2) = -6. Substitute this value of y in (1) to get:

At x=2, (dy/dt) = -(5/6) is the rate of change of curve with respect time.

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