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
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.)
4 answers
Steve is tha goat
Thanks bestie Steve! Hope ur thriving bc im not! <3
so true
1 answer