Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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.

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.