# 4.1 Related rates  (Page 4/7)

 Page 4 / 7

## Water draining from a funnel

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. Water is draining from a funnel of height 2 ft and radius 1 ft. The height of the water and the radius of water are changing over time. We denote these quantities with the variables h and r , respectively.

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 $\frac{dV}{dt}=-0.03\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$

Step 3: The volume of water in the cone is

$V=\frac{1}{3}\pi {r}^{2}h.$

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

$V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}\phantom{\rule{0.1em}{0ex}}{h}^{3}.$

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time $t,$ we obtain

$\frac{dV}{dt}=\frac{\pi }{4}\phantom{\rule{0.1em}{0ex}}{h}^{2}\frac{dh}{dt}.$

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 $\frac{dV}{dt}=-0.03{\phantom{\rule{0.2em}{0ex}}\text{ft}}^{3}\text{/sec}.$ Therefore,

$-0.03=\frac{\pi }{4}{\left(\frac{1}{2}\right)}^{2}\frac{dh}{dt}\text{},$

which implies

$-0.03=\frac{\pi }{16}\phantom{\rule{0.2em}{0ex}}\frac{dh}{dt}.$

It follows that

$\frac{dh}{dt}=-\frac{0.48}{\pi }=-0.153\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$

At what rate is the height of the water changing when the height of the water is $\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{ft}?$

$-0.61\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$

## Key concepts

• To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time.
• In terms of the quantities, state the information given and the rate to be found.
• Find an equation relating the quantities.
• Use differentiation, applying the chain rule as necessary, to find an equation that relates the rates.
• Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.

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 $x=-2$ and $y=2{x}^{2}+1$ if $\frac{dy}{dt}=-1.$

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, $\text{Ω}\right)$ 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 $0.5\phantom{\rule{0.2em}{0ex}}\text{Ω}\text{/}\text{min}$ and ${R}_{2}$ decreases at a rate of $1.1\text{Ω/min},$ at what rate does the total resistance change when ${R}_{1}=20\text{Ω}$ and ${R}_{2}=50\text{Ω}\text{/}\text{min}?$

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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