1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dV/dt, the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/dt, the change in the radius with respect to time. Remember that the volume of a sphere is V=4/3 pi r^3.
2. A baseball diamond is a square 90 feet on a side. A player runs from first base to second base at a rate of 15 feet per second. At what rate is the player's distance from third base decreasing when the player is halfway between first and second base? We've already set part of this problem up. If we let x be the distance between the player and second base,and y be the distance between the player and third base, then dx/dt=-15 feet per second, and dy/dt will tell us what we want to know. Use the picture to find a relationship that will help you answer the question.
3. A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per second, and we're looking for dv/dt . Use the picture to help you find the relationship between x and y, and use it to answer the question asked here.
4. Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = pi(r^2)h, where r is the radius and h is the height of this cylinder.)
These are all just applications of the chain rule. If y is a function of u, and u is a function of t, then
dy/dt = dy/du * du/dt
So, now we see what do do:
#1
v=4/3 π r^3
dv/dt = 4πr^2 dr/dt
dr/dt = 15/(4πr^2)
As expected, if the volume changes at a steady rate, the bigger the balloon gets, the slower the radius increases.
#2
Using the Pythagorean Theorem, we know that
x^2 + 90^2 = y^2
x dx/dt = y dy/dt
So, find y when x=45, and then just plug in dx/dt = -15
#3
You don't say what y is, but suppose y is the distance of the tip of the shadow from the man. In that case, using similar triangles,
y/2 = (x+y)/5
y = 2/3 x
So,
dy/dt = 2/3 dx/dt
#4
1st way: use the product rule
v = πr^2h
dv/dt = 2πrh dr/dt + πr^2 dh/dt
You know that dv/dt=0, since the volumem of oil is constant.
Now, what is h? πr^2h = 1, so h = 1/(π*.01^2)
Now you can plug in the numbers.
2nd way: use the fact that v is constant
πr^2h = 1, so
πh = 1/r^2
π dh/dt = -2/r^3 dr/dt
now just plug in your numbers to find dh/dt
don't forget to watch the units.
1m^3 = 10^6 cm^3
Steve is tha goat
Thanks bestie Steve! Hope ur thriving bc im not! <3
so true
1. The volume of a soap bubble is changing, but the change in the volume doesn't seem to be so bubbly. We need to calculate the rate at which the radius changes. Remember the equation for the volume of a sphere: V = 4/3 πr^3. We're given dV/dt = 15 cm^3/s, but we're interested in dr/dt. So, we need to differentiate this equation with respect to time. But beware of the pi, it might be irrational but it still has feelings too!
2. Ah, the baseball diamond problem. A player running from first to second base at 15 feet per second, while we're interested in the rate at which the player's distance from third base is *decreasing*. It's like a game of "Catch (up) me if you can!" We're given dx/dt = -15 ft/s, and we need to know dy/dt. Don't worry, I won't make any baseless jokes here!
3. Now this is a shadowy problem! A man walking towards a streetlight, casting a shadow that's moving with him. The man is 2 meters tall, walking at a pace of 2 meters per second, and the streetlight is 5 meters high. We're asked about the rate at which the tip of the shadow is moving. Remember, dx/dt = -2 m/s, but we want to find dv/dt. Time to throw some light on this!
4. Ah, the oil slick situation. A circular oil slick that's spreading, while the thickness of the oil is decreasing at a snail's pace of 0.1 cm/hr. We're interested in the rate at which the radius is increasing. The volume of this slick can be calculated using V = πr^2h, where r is the radius and h is the height. We're given dV/dt = -0.1 cm/hr, but we need to figure out dr/dt. It's a slippery problem, but we'll get through it!
1. To find the rate of change of the radius when blowing air into a soap bubble, we need to relate the change in volume to the change in radius. We're given dV/dt = 15 cubic centimeters per second, and we want to find dr/dt.
The volume of a sphere is given by V = 4/3 * π * r^3. So, we can express the volume as a function of the radius: V(r) = 4/3 * π * r^3.
To find the rate of change of the volume with respect to time, we differentiate the equation V(r) = 4/3 * π * r^3 with respect to time (t) using the chain rule:
dV/dt = dV/dr * dr/dt.
We already know dV/dt = 15 cubic centimeters per second (given in the question).
Now, let's differentiate the equation V(r) = 4/3 * π * r^3 with respect to r:
dV/dr = 4/3 * π * 3r^2 = 4πr^2.
Now we can substitute the known values:
15 = (4πr^2) * dr/dt.
Simplifying the equation, we can solve for dr/dt:
dr/dt = 15 / (4πr^2).
2. In this problem, we're given that the player runs from first base to second base at a rate of 15 feet per second. We want to find the rate at which the player's distance from third base is decreasing when the player is halfway between first and second base.
Let's set up the problem using x and y as defined in the question. x represents the distance between the player and second base, and y represents the distance between the player and third base.
We are given dx/dt = -15 feet per second (negative because the player is moving away from first base).
To relate dx/dt and dy/dt, we can use similar triangles. The triangle formed by the player, first base, and second base is similar to the triangle formed by the player, second base, and third base.
Since both triangles are similar, their corresponding sides are proportional. Therefore:
x/y = (90 - x)/y.
Now we differentiate both sides of the equation with respect to time (t):
(dx/dt)/y - x(dy/dt)/y^2 = 0.
We can simplify and solve for dy/dt:
(dy/dt) = (dx/dt) * (y/x) = (-15) * (y/x).
3. In this problem, a man 2 meters tall walks towards a streetlight that is 5 meters above the ground. We're asked to find the rate at which the tip of his shadow is moving. Let's define x as the horizontal distance from the man to the streetlight, and y as the length of his shadow.
We're given dx/dt = -2 meters per second (negative because the man is moving closer to the streetlight).
To relate dx/dt and dy/dt, we can use similar triangles. The triangle formed by the man, his shadow, and the streetlight is similar to the triangle formed by the man, the streetlight, and the top of the shadow.
Since both triangles are similar, their corresponding sides are proportional. Therefore:
x/y = (x + 5)/2.
Now we differentiate both sides of the equation with respect to time (t):
(dx/dt)/y - x(dy/dt)/y^2 = 0.
We can simplify and solve for dy/dt:
(dy/dt) = (dx/dt) * (y/x) = (-2) * (y/x).
4. In this problem, a circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at a rate of 0.1 cm/hr as the slick spreads. We're asked to find the rate at which the radius of the slick is increasing when the radius is 8 meters.
Let's define V as the volume of the oil slick, r as the radius, and h as the height (thickness) of the oil slick.
We're given dh/dt = -0.1 cm/hr (negative because the thickness is decreasing).
The volume of the oil slick can be expressed as V = π * r^2 * h. Since the slick is very flat, we can consider it as a cylinder.
To find the rate at which the radius is increasing (dr/dt), we'll differentiate the volume equation with respect to time (t):
dV/dt = dV/dr * dr/dt + dV/dh * dh/dt.
Since the slick is spreading in a circular shape, the change in volume with respect to height (dV/dh) is equal to the area of the base:
dV/dh = π * r^2.
Since the height (h) is decreasing, dh/dt is negative (-0.1 cm/hr).
Now, let's evaluate dV/dt when the radius is 8 meters:
dV/dt = π * (8^2) * (-0.1) = -64π * 0.1 = -6.4π cubic meters per hour.
We're interested in finding dr/dt, so we rearrange the equation:
dV/dt = dV/dr * dr/dt + dV/dh * dh/dt.
-6.4π = (π * r^2) * dr/dt + (π * r^2) * (-0.1).
Simplifying the equation, we can solve for dr/dt:
dr/dt = (-6.4 + 0.1) / (π * r^2).
dr/dt = -6.3 / (π * r^2).