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
1 answer:
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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