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Water is draining from the bottom of a cone-shaped funnel at the rate of $0.0{3\phantom{\rule{0.2em}{0ex}}\text{ft}}^{3}\text{/sec}.$ The height of the funnel is 2 ft and the radius at the top of the funnel is $1\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At what rate is the height of the water in the funnel changing when the height of the water is $\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{ft}?$
Step 1: Draw a picture introducing the variables.
Let $h$ denote the height of the water in the funnel, $r$ denote the radius of the water at its surface, and $V$ denote the volume of the water.
Step 2: We need to determine $\frac{dh}{dt}$ when $h=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{ft}.$ We know that $\frac{dV}{dt}=\mathrm{-0.03}\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$
Step 3: The volume of water in the cone is
From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, $\frac{r}{h}=\frac{1}{2}$ or $r=\frac{h}{2}.$ Using this fact, the equation for volume can be simplified to
Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time $t,$ we obtain
Step 5: We want to find $\frac{dh}{dt}$ when $h=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{ft}.$ Since water is leaving at the rate of $0.0{3\phantom{\rule{0.2em}{0ex}}\text{ft}}^{3}\text{/sec},$ we know that $\frac{dV}{dt}=\mathrm{-0.03}{\phantom{\rule{0.2em}{0ex}}\text{ft}}^{3}\text{/sec}.$ Therefore,
which implies
It follows that
At what rate is the height of the water changing when the height of the water is $\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{ft}?$
$\mathrm{-0.61}\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$
For the following exercises, find the quantities for the given equation.
Find $\frac{dy}{dt}$ at $x=1$ and $y={x}^{2}+3$ if $\frac{dx}{dt}=4.$
$8$
Find $\frac{dx}{dt}$ at $x=\mathrm{-2}$ and $y=2{x}^{2}+1$ if $\frac{dy}{dt}=\mathrm{-1}.$
Find $\frac{dz}{dt}$ at $\left(x,y\right)=\left(1,3\right)$ and ${z}^{2}={x}^{2}+{y}^{2}$ if $\frac{dx}{dt}=4$ and $\frac{dy}{dt}=3.$
$\frac{13}{\sqrt{10}}$
For the following exercises, sketch the situation if necessary and used related rates to solve for the quantities.
[T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, $\text{\Omega})$ is given by the equation $\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}.$ If ${R}_{1}$ is increasing at a rate of $0.5\phantom{\rule{0.2em}{0ex}}\text{\Omega}\text{/}\text{min}$ and ${R}_{2}$ decreases at a rate of $1.1\text{\Omega /min},$ at what rate does the total resistance change when ${R}_{1}=20\text{\Omega}$ and ${R}_{2}=50\text{\Omega}\text{/}\text{min}?$
A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?
$2\sqrt{3}$ ft/sec
A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially $20\phantom{\rule{0.2em}{0ex}}\text{ft}$ away from the wall, how fast does the ladder move up the wall $5\phantom{\rule{0.2em}{0ex}}\text{sec}$ after we start pushing?
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