Question

Suppose that x=x(t) and y=y(t) are both functions of t. If
y^2+x=18,
and dy/dt=−4 when x=2 and y=−4, what is dx/dt?

Answer 1

\[\frac{ dx }{ dt }=\frac{ dx }{ dy }\frac{ dy }{ dt }\]

Answer 2

iv gotten as far as dv/dt=(3pi/12)h^2*(dh/dt)

Answer 3

\[x=18-y^2\]
\[\frac{ dx }{ dy }=-2y\]but y=-4
\[\frac{ dx }{ dt }=(-4)\times -2(-4)\]

Answer 4

thanks dont know what i did but had -1/32

Answer 5

probably differenciated y with respect to x,

Answer 6

probably

Answer 7

can you help me with this one?
Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 23 feet high?

Answer 8

volume of a right circular cone with height h and radius of the base r is given by v=(1/3)pi(r^2)h

Answer 9

\[\frac{ dV }{ dt }=10\]
\[\frac{ d V}{ dh }=\pi(1/4) h^2\]

Answer 10

\[\frac{ dh }{ dt }=\frac{ dV }{ dt }\frac{ dh }{ dV }\]
\[2r=h,r=\frac{ h }{ 2 }\]
\[V=\frac{ 1 }{ 3 }\frac{ 1 }{ 4 }h^3=\frac{ 1 }{ 12 }h^3\]

Answer 11

\[\frac{ dh }{ dV }=\frac{ 4 }{ \pi h^2 }\]by swapping the above derivative
so\[\frac{ dh }{ dt }=(10)(\frac{ 4 }{ \pi(h^2) })=\frac{ 40 }{ 23^2\pi }\]

Answer 12

@NickR

Answer 13

oh ok, that helped alot. thank you for your help

Answer 14

wish i could give you another medel