A point moves along the curve #y=2x^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=-3#?

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Answer 1 · 2018-04-05T13:57:21

# 1/6"(unit)/sec."#

We are given that #y=2x^2+1#.

Diff.ing w.r.t. #t# using the Chain Rule, we get,

#dy/dt=d/dt{2x^2+1}=d/dx{2x^2+1}*dx/dt, i.e., #

# dy/dt=4x*dx/dt...........(star)#.

But, here,

#dy/dt="the rate of change of "y, and,#

#dx/dt"=the reqd. rate of change of "x.#

If we use #-ve# sign to indicate decreasing rate, then,

by what is given, #dy/dt=-2"(unit)/sec.", x=-3, dx/dt=?#.

#:. (star) rArr -2=4(-3)*dx/dt, or, dx/dt=1/6"(unit)/sec."#