There are two values depending on the point.
##{dx}/{dt}={(-8 ” at ” (3,4)),(8 ” at “(-3,4)):}##
Let us look at some details.
First, let us find the values of ##x##.
By plugging in ##y=4## into ##x^2+y^2=25##,
##x^2+16=25 Rightarrow x^2=9 Rightarrow x=pm3##
Now, let us find some derivatives.
By differentiating with respect to ##t##,
##d/{dt}(x^2+y^2)=d/{dt}(25) Rightarrow 2x{dx}/{dt}+2y{dy}/{dt}=0##
by dividing by ##2x##,
##Rightarrow {dx}/{dx}+y/x{dy}/{dt}=0##
by subtracting ##y/x{dy}/{dt}##,
##Rightarrow {dx}/{dt}=-y/x{dy}/{dt}##
Since ##y=4##, ##x=pm3##, and ##{dy}/{dt}=6##,
##{dx}/{dt}=-{4}/{pm3}(6)=pm8##
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Get The Answer If ##x^2+y^2=25## and ##dy/dt=6##, how do you find ##dx/dt## when ##y=4## ?
There are two values depending on the point.
##{dx}/{dt}={(-8 ” at ” (3,4)),(8 ” at “(-3,4)):}##
Let us look at some details.
First, let us find the values of ##x##.
By plugging in ##y=4## into ##x^2+y^2=25##,
##x^2+16=25 Rightarrow x^2=9 Rightarrow x=pm3##
Now, let us find some derivatives.
By differentiating with respect to ##t##,
##d/{dt}(x^2+y^2)=d/{dt}(25) Rightarrow 2x{dx}/{dt}+2y{dy}/{dt}=0##
by dividing by ##2x##,
##Rightarrow {dx}/{dx}+y/x{dy}/{dt}=0##
by subtracting ##y/x{dy}/{dt}##,
##Rightarrow {dx}/{dt}=-y/x{dy}/{dt}##
Since ##y=4##, ##x=pm3##, and ##{dy}/{dt}=6##,
##{dx}/{dt}=-{4}/{pm3}(6)=pm8##
Previous answers to this question
This is a preview of an assignment submitted on our website by a student. If you need help with this question or any assignment help, click on the order button below and get started. We guarantee authentic, quality, 100% plagiarism free work or your money back.
Get The Answer