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A particle moves along the curve y = 5x^2 – 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = 1?


Decreasing one fifth unit/sec

Increasing one fifth unit/sec

Decreasing one tenth unit/sec

Increasing one tenth unit/sec

Sagot :

Answer:

The correct option is;

Increasing one fifth unit/sec

Step-by-step explanation:

The equation that gives the curve of the particle of the particle is y = 5·x² - 1

The rate of decrease of the y value dy/dt = 2 units per second

We have;

dy/dx = dy/dt × dt/dx

dy/dx = 10·x

dy/dt = 2 units/sec

dt/dx = (dy/dx)/(dy/dt)

dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)

When x = 1

dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec

dx/dt = 1/5 unit/sec

Therefore, x is increasing one fifth unit/sec.

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