Ask your own question, for FREE!
Mathematics 24 Online
OpenStudy (anonymous):

A particle moves along the curve y = 5x2 – 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?

OpenStudy (anonymous):

@@welshfella

OpenStudy (anonymous):

@welshfella

OpenStudy (anonymous):

@ganeshie8 @Destinymasha @e.mccormick @EclipsedStar @Directrix plz help :)

OpenStudy (ribhu):

at what rate y is changing i guess ur question needs correction @mathrulezz

OpenStudy (anonymous):

haha sorry i think they posted it on my website wrong :) but yes i guess that's what they mean :)

OpenStudy (jhannybean):

rate of change = slope of the given function = derivative.

OpenStudy (anonymous):

okay so i have 2 find the derivatiev ? :)

OpenStudy (jhannybean):

at each time interval, you're going to have a different rate of change.

OpenStudy (anonymous):

10x is the derivative :D

OpenStudy (jhannybean):

Yep, you're finding the derivative.

OpenStudy (anonymous):

but it asks at what rate is x changing when x = 1

OpenStudy (jhannybean):

y' = 10x , what is x when y' = 2 units/sec?

OpenStudy (anonymous):

1/5 :) thanku!

OpenStudy (anonymous):

so its decreasing?

OpenStudy (jhannybean):

That's what your problem says :)

OpenStudy (anonymous):

graCIAS! thank u so much

OpenStudy (jhannybean):

No problem

OpenStudy (ribhu):

y = 5x^2 dy/dt = 10xdx/dt; rate of change of y is given and you have to calculate rate of change of x at x=1. dy/dt = -2 since it is decreasing. so dx/dt = -1/5. change of rate of x

OpenStudy (ribhu):

@Jhannybean i am also correct with this one.

OpenStudy (jhannybean):

There you go, you are right.

OpenStudy (jhannybean):

Because y and x are both changing across a curve with respect to time.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!