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A ladder of length $L$ feet leans against a vertical wall (assume that the floor is horizontal). Suppose that the base of the ladder is sliding away from the wall at a rate of 4 ft/s. How fast is the top of the ladder falling when the base of the ladder is $L/2$ feet from the wall?
If $x$ is the distance from the wall along the floor, and $y$ is the height of the top of the ladder, we know $x^2 + y^2 = L^2$ from the Pythagorean theorem. The final answer is $-4/\sqrt{3}$ ft/s.
Seven-foot-tall Yogi Bear watches, from twenty feet away, a balloon carry away a picnic basket. If the picnic basket rises straight up at a constant rate of 6 ft/s, how fast is the angle of Yogi’s head turning when the picnic basket is 27 feet above the ground?
Let $y$ be the height of the picnic basket above seven feet (which is Yogi Bear's eyeline), and $\theta$ be the angle of inclination that his head makes. This gives us two legs of a right triangle, so $\tan \theta = x/20$. The final answer is $3/20$ rad/sec.
Helium is being pumped into a spherical balloon at a rate of 20 cm3/s. At what rate is the surface area increasing when the radius is 10 cm? What happens to the rate as the balloon grows larger and larger?
Since the surface area relates to the radius, you can first find how the radius changes as the volume changes. Using the formula for the volume of a sphere, one arises at
$$\frac{dr}{dt} = \frac{1}{4\pi r^2} \frac{dV}{dt}$$ Of course, the surface area of a sphere is given by $S = 4\pi r^2$, so ultimately, $$\frac{dS}{dt} = 4 \textrm{ cm}^3/\textrm{s}$$As $r \rightarrow \infty$, the rate approaches $0$.
A particle moves along the parabola $y=x^2$ in the first quadrant, in such a way that its $x$-coordinate increases at a constant 10 units/s. How fast is the angle of inclination of the line joining the particle to the origin changing when $x=3$? How fast is the angle of inclination of the line tangent to the parabola at the particle changing when $x = \sqrt{3}/2$?
The tangent of the angle of inclination of a line is, by definition, its slope (i.e., rise over run). In the first case, the line goes through the origin and $(x,x^2)$, so its slope is $x$. This tells us that $\tan \theta = x$. The final answer is $\frac{d\theta}{dt} = 1$ rad/s. In the second case, the slope of the tangent line is the derivative of the function, $2x$. This makes the equation in question $\tan \theta = 2x$, which leads to a final answer of $5$ rad/s.
Let $R$ be a rectangle whose sides are changing size. If the length increases at a rate of 2 inches per minute and the width is decreasing at a rate of 1.5 inches per minute, what is the rate of change of the area when the length is 30 inches and the width is 20 inches? If the area increases at a rate of 150 square inches per minute and the length increases at a rate of 3 inches per minute, what is the rate of change of the width when the length is 25 inches and the width is 20 inches?
Both questions use the formula for the area of a rectangle (i.e., $A = w \cdot l$). Remembering to use the product rule when taking the appropriate derivative, we find $\frac{dA}{dt} = -5$ in2/min in the first case and $\frac{dw}{dt} = 18/5$ in/min in the second case.
Determine the points in the first quadrant where the tangent line to $x^3+y^3=3xy$ is horizontal or vertical.
We hope to find points $(x,y)$ with $x,y \gt 0$ (so they are in the first quadrant), where $\frac{dy}{dx} = 0$ (the points where there is a horizontal tangent) or where $\frac{dx}{dy} = 0$ (the points where there is a vertical tangent).
Differentiating the given equation with respect to $x$ to introduce $\frac{dy}{dx}$, we find
$$3x^2 + 3y^2 \cdot \textstyle{\frac{dy}{dx}} = 3 (x \cdot \textstyle{\frac{dy}{dx}} + y)$$Solving for $\frac{dy}{dx}$, we see that
$$\begin{array}{rcl} x^2+y^2 \cdot \frac{dy}{dx} &=& x \cdot \frac{dy}{dx} + y\\ (y^2-x) \cdot \frac{dy}{dx} &=& y - x^2\\ \frac{dy}{dx} &=& \displaystyle{\frac{y-x^2}{y^2-x}} \end{array}$$So when $\frac{dy}{dx} = 0$ it must be that $y-x^2=0$, or equivalently, that $y=x^2$.
Recall we were given that $x^3 + y^3 = 3xy$, so substituting $y=x^2$ we find
$$x^3 + (x^2)^3 = 3x \cdot x^2$$This simplifies to
$$x^3(x^3-2) = 0$$suggesting that $x = 0$ or $x = \sqrt[3]{2}$.
Using $y=x^2$, we have the corresponding $y$-values of $0$ and $\sqrt[3]{4}$.
Thus, the point where we have a horizontal tangent in the first quadrant is given by $(\sqrt[3]{2},\sqrt[3]{4})$.
By symmetry, we can solve for where $\frac{dx}{dy}=0$ to find this happens in the first quadrant at $(\sqrt[3]{4},\sqrt[3]{2})$.
The length of a rectangle is increasing at the rate of $3$ inches per minute, while the width is decreasing at the rate of 2 inches per minute. Find the rate of change of the area of the rectangle when the length is 5 inches and the width is 3 inches.
$-1$ in2/min
A boat is pulled in by means of a winch on a dock 12 ft above the deck of a boat. If the winch pulls in rope at the rate of 4 ft/sec, determine the speed of the boat when 13 ft of rope is out.
$\displaystyle{\frac{52}{5}}$ ft/sec
The radius of a right circular cylinder is increasing at the rate of 3 cm/sec. If the volume of the cylinder remains constant, find the rate at which the height of the cylinder is changing when the radius is 5 cm and the height is 4 cm.
$\displaystyle{\frac{-24}{5}}$ cm/sec
A conical tank (vertex down) is 10 ft across the top and 12 ft deep. If water is flowing into the tank at the rate of 10 ft$^3$/min, find the rate of change of the depth of water the instant it is 8 feet deep.
$\displaystyle{\frac{9}{10\pi}}$ ft/min
A weather balloon is released at 9:00 a.m. and rises vertically at the rate of 25 ft/min. An observer stands on level ground 400 feet from the balloon's release point. How fast is the distance between the balloon and the observer's feet changing at 9:12 a.m.?
$15$ ft/min
Water is flowing into a cylindrical tank of radius 2 ft at the rate of 8 ft$^3$/min. How fast is the water level rising?
$\displaystyle{\frac{2}{\pi}}$ ft/min
The volume of a right circular cone is increasing at the rate of $25\pi$ in$^3$ per minute and the diameter of the base of the cone is decreasing at the rate of 1 inch per minute. At a certain instant the base diameter is 12 inches long and the height is 18 inches. What is the rate of change in the height of the cone at this instant?
$\displaystyle{\frac{61}{12}}$ in/min
The side of an equilateral triangle is $2\sqrt{3}$ inches long and is increasing at the rate of 4 inches per second. Find the rate of change in the area of the triangle at this moment.
$12$ in2/sec
Water is poured into a cone at the rate of 12 ft$^3$ per minute. If the cone is 12 feet high and 6 feet in diameter, how fast is the water level rising when the water is 7 feet deep? (The vertex of the cone is down.)
$\displaystyle{\frac{192}{49\pi}}$ ft/sec
A spherical balloon is expanding under the influence of solar radiation. If its radius is increasing at the rate of 2 in/min, how fast is the volume increasing when the radius is 5 inches?
$200\pi$ in3/min
A streat urchin is sitting at the base of a wall 5 feet high. He is holding one end of a string; a wharf rat is on the other end of the string. As the rat runs along the top of the wall, the urchin lets out string at the rate of 2 feet per second, but the string remains taut. Find the rate at which the rat is moving along the wall when 13 feet of string has been let out by the urchin. How does your answer change if the rat is running along the base of the wall?
$\displaystyle{\frac{13}{6}}$ ft/sec along the top of the wall; $2$ ft/sec along the base of the wall.
The base of a triangle is increasing at the rate of 3 inches per minute while the altitude is decreasing at the same rate. At what rate is the area changing when the base is 10 inches wide and the altitude is 6 inches high?
$-6$ in2/min
A spotlight is on the ground 100 feet from a building that has vertical sides. A person 6 feet tall starts at the spotlight and walks toward the building at a rate of 5 feet per second. How fast is the top of the shadow moving \underline{down} the building when the person is 50 feet away from the building?
$\displaystyle{\frac{6}{5}}$ ft/sec
The volume of a cylinder is increasing at the rate of $48\pi$ in$^3$ per second. The height of the cylinder is always twice the radius. Find the rate of change in the surface area of the cylinder when the volume of the cylinder is $128\pi$ in$^3$.
$24\pi$ in2/sec
A conical paper cup (vertex down) is leaking water at the rate of $4\pi$ in$^3$ per minute. If the cup is 12 inches high and 6 inches in diameter, at what rate is the water level being lowered at the instant the top surface area of the water is $4\pi$?
$-1$ in/min