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The distance is increasing at a rate of $62.50$ ft/sec.

The distance is decreasing at a rate of $11.99$ ft/sec.

T

F

(a) $5/(2\pi) \approx 0.796$cm/s (b) $1/(4\pi)\approx 0.0796$ cm/s (c) $1/(40\pi)\approx 0.00796$ cm/s

(a) $5/(2\pi)\approx 0.796$cm/s (b) $1/(40\pi)\approx 0.00796$ cm/s (c) $1/(4000\pi)\approx 0.0000796$ cm/s

Due to the height of the plane, the gun does not have to rotate very fast.

(a) $0.0573$ rad/s (b) $0.0725$ rad/s (c) In the limit, rate goes to $0.0733$ rad/s

(a) $0.04$ ft/s (b) $0.458$ ft/s (c) $3.35$ ft/s (d) Not defined; as the distance approaches 24, the rates approaches $\infty$.

(a) $30.59$ ft/min (b) $36.1$ ft/min (c) $301$ ft/min (d) The boat no longer floats as usual, but is being pulled up by the winch (assuming it has the power to do so).

(a) $50.92$ ft/min (b) $0.509$ ft/min (c) $0.141$ ft/min

As the tank holds about 523.6ft$^3$, it will take about 52.36 minutes.

(a) $0.63$ ft/sec (b) $1.6$ ft/sec

About 52 ft.

(a) The balloon is 105ft in the air. (b) The balloon is rising at a rate of 17.45ft/min. (Hint: convert all angles to radians.)

The cone is rising at a rate of $0.003$ft/s.

We have an equation relating $P$ and $Q$:

$$\begin{aligned} P&=Q^3 \\ \text{We differentiate implicitly with respect to a third variable, $t$:} \\ \frac{\mathrm{d}P}{\mathrm{d}t}&=3Q^2\cdot\frac{\mathrm{d}Q}{\mathrm{d}t} \end{aligned}$$

If we know two of the three quantities $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}$, $Q$, and $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}$, then we can find the third. Therefore, ii is a question we can solve. If we know $P$, then we also know $Q$ (it's just the cube root of $P$), so also we can solve iv. However, if we know neither $P$ nor $Q$, then we can't find $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}$ based only off $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}$, and we can't find $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}$ based only off $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}$. So we can't solve i or iii.

As is so often the case, we use a right triangle in this problem to relate the quantities.

$$\begin{aligned} \sin\theta&=\frac{D}{2}\\ D&=2\sin\theta\\ \text{Using the chain rule, we differentiate both sides with respect to time, $t$.} \\ \frac{\mathrm{d}D}{\mathrm{d}t}&=2\cos\theta\cdot\frac{\mathrm{d}\theta}{\mathrm{d}t} \\ \text{So, if $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}=0.25$ radians per hour and $\theta = \dfrac{\pi}{4}$ radians, then} \\ (a)\qquad\frac{\mathrm{d}D}{\mathrm{d}t}&=2\cos\left(\frac{\pi}{4}\right)\cdot 0.25=2\left(\frac{1}{\sqrt{2}}\right)\frac{1}{4}=\frac{1}{2\sqrt{2}}~\text{metres per hour}. \end{aligned}$$

Setting aside part (b) for a moment, let's think about (c). If $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$ and $\displaystyle \frac{\mathrm{d}D}{\mathrm{d}t}$ have different signs, then because $\displaystyle \frac{\mathrm{d}D}{\mathrm{d}t}=2\cos\theta\cdot\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$, that means $\cos\theta<0$. We have to have a nonnegative depth, so $D>0$ and $D=2\sin\theta$ implies $\sin \theta >0$. If $\sin\theta\ge0$ and $\cos \theta<0$, then $\theta\in(\pi/2,\pi]$. On the diagram, that looks like this:

That is: the water has reversed direction. This happens, for instance, when a river empties into the ocean and the tide is high. Skookumchuck Narrows provincial park, in the Sunshine Coast, has reversing rapids.

Now, let's return to (b). If the rope is only 2 metres long, and the river rises higher than 2 metres, then our equation $D=2\sin\theta$ doesn't work any more: the buoy might be stationary underwater while the water rises or falls (but stays at or above $2$ metres deep).

(a) When the point is at $(0,-2)$, its $y$-coordinate is not changing, because it is moving along a horizontal line. So, the rate at which the particle moves is simply $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}$. Let $\theta$ be the angle an observer would be looking at, in order to watch the point. Since we know $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$, a reasonable plan is to find an equation relating $\theta$ and $x$, and then differentiate implicitly with respect to $t$. To do this, let's return to our diagram.

When the point is a little to the right of $(0,-2)$, then we can make a triangle with the origin, as shown. If we let $\theta$ be the indicated angle, then $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}=1$ radian per second. (It is given that the observer is turning one radian per second, so this is how fast $\theta$ is increasing.) From the right triangle in the diagram, we see $$\tan\theta=\frac{x}{2}$$ Now, we have to take care of a subtle point. The diagram we drew only makes sense for the point when it is at a position a little to the right of $(0,-2)$. So, right now, we've only made a set-up that will find the derivative from the right. But, with a little more thought, we see that even when $x$ is negative (that is, when the point is a little to the left of $(0,-2)$), our equation holds if we are careful about how we define $\theta$. Let $\theta$ be the angle between the line connecting the point and the origin, and the $y$-axis, where $\theta$ is negative when the point is to the left of the $y$-axis.

Since $x$ and $\theta$ are both negative when the point is to the left of the $y$-axis,

$$\begin{aligned} \tan|\theta|&=\frac{2}{|x|}\\ \tan(-\theta) &= \frac{-x}{2} \\ \text{So, since $\tan(-\theta)=-\tan(\theta)$:} \\ \tan\theta&=\frac{x}{2} \end{aligned}$$

So, we've shown that the relationship $\tan\theta=\dfrac{x}{2}$ holds when our point is at $(x,-2)$, regardless of the sign of $x$.

Moving on, since we are given $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}$ and asked for $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}$, we differentiate with respect to $t$:

$$\begin{aligned} \sec^2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t}&=\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t} \\ \text{When the point is at $(0,-2)$, since the observer is turning at one radian per second, also $\displaystyle \frac{\mathrm{d}\theta}{\mathrm{d}t}=1$. Also, looking at the diagram, $\theta=0$. Plugging in these values:} \\ \sec^2\left(0\right)\cdot(1)&=\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}\\ 1&=\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}\\ \frac{\mathrm{d}x}{\mathrm{d}t}&=2 \end{aligned}$$

So, the particle is moving at 2 units per second.

(b) When the point is at $(0,2)$, it is moving along a line with slope $-\frac{1}{2}$ and $y$-intercept $2$. So, it is on the line $$y=2-\frac{1}{2}x$$ That is, at time $t$, if the point is at $(x(t),y(t))$, then $x(t)$ and $y(t)$ satisfy $y(t)=2-\frac{1}{2}x(t)$. Implicitly differentiating with respect to $t$: $$\frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{1}{2}\cdot\frac{\mathrm{d}x}{\mathrm{d}t}$$ So, when $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=1$, $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=-\dfrac{1}{2}$. That is, its $y$-coordinate is decreasing at $\dfrac{1}{2}$ unit per second.

For the question “How fast is the point moving?", remember that the velocity of an object can be found by differentiating (with respect to time) the equation that gives the position of the object. The complicating factors in this case are that (1) the position of our object is not given as a function of time, and (2) the position of our object is given in two dimensions (an $x$ coordinate and a $y$ coordinate), not one.

Remark: the solution below is actually pretty complicated. It is within your abilities to figure it out, but later on in your mathematical career you will learn an easier way, using vectors. For now, take this as a relatively tough exercise, and a motivation to keep learning: your intuition that there must be an easier way is well founded!

The point is moving along a straight line. So, to take care of complication (2), we can give its position as a point on the line. We can take the line as a sort of axis. We'll need to choose a point on the axis to be the “origin": $(2,1)$ is a convenient point. Let $D$ be the point's (signed) distance along the “axis" from $(2,1)$. When the point is a distance of one unit to the left of $(2,1)$, we'll have $D=-1$, and when the point is a distance of one unit to the right of $(2,1)$, we'll have $D=1$. Then $D$ changes with respect to time, and $\displaystyle \frac{\mathrm{d}D}{\mathrm{d}t}$ is the velocity of the point. Since we know $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}$ and $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}$, a reasonable plan is to find an equation relating $x$, $y$, and $D$, and differentiate implicitly with respect to $t$. (This implicit differentiation takes care of complication (1).) Using the Pythagorean Theorem:

$$\begin{aligned} D^2&=(x-2)^2+(y-1)^2 \\ \text{Differentiating with respect to $t$:} \\ 2D\cdot\frac{\mathrm{d}D}{\mathrm{d}t}&=2(x-2)\cdot\frac{\mathrm{d}x}{\mathrm{d}t}+2(y-1)\cdot\frac{\mathrm{d}y}{\mathrm{d}t} \\ \text{We plug in $x=0$, $y=2$, $\frac{\mathrm{d}x}{\mathrm{d}t}=1$, $\frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{1}{2}$, and $D=-\sqrt{(0-2)^2+(2-1)^2}=-\sqrt{5}$ (negative because the point is to the left of $(2,1)$):} \\ -2\sqrt{5}\cdot\frac{\mathrm{d}D}{\mathrm{d}t}&=2(-2)(1)+2(1)\left(-\frac{1}{2}\right)\\ \frac{\mathrm{d}D}{\mathrm{d}t}&=\frac{\sqrt{5}}{2}~\text{units per second} \end{aligned}$$

Since $A=0$, the equation relating the variables tells us:

$$\begin{aligned} 0&=\log\left(C^2+D^2+1\right)\\ 1&=C^2+D^2+1\\ 0&=C^2+D^2\\ 0&=C=D \\ \text{This will probably be useful information. Since we're also given the value of a derivative, let's differentiate the equation relating the variables implicitly with respect to $t$. For ease of notation, we will write $\displaystyle \frac{\mathrm{d}A}{\mathrm{d}t}=A'$, etc.} \\ A'B+AB'&=\frac{2CC'+2DD'}{C^2+D^2+1} \\ \text{At $t=10$, $A=C=D=0$:} \\ A'B+0&=\frac{0+0}{0+0+1}\\ A'B&=0 \\ \text{at $t=10$, $A'=2$ units per second:} \\ 2B&=0\\ B&=0. \end{aligned}$$