related rates Archives - rweber.net https://www.rweber.net/tag/related-rates/ trying to be a mile wide AND a mile deep Fri, 04 Dec 2015 01:09:43 +0000 en-US hourly 1 https://wordpress.org/?v=6.9.4 37896774 Distinguishing related rates and optimization https://www.rweber.net/mathematics/calculus/distinguishing-related-rates-and-optimization/ https://www.rweber.net/mathematics/calculus/distinguishing-related-rates-and-optimization/#comments Mon, 11 Mar 2013 12:00:15 +0000 http://www.rweber.net/?p=272 One of the main things students struggle with on exams is identifying the method to use on each problem. When they are learned in class, they are segregated by type, and their distinguishing features are not always highlighted. Here’s a brief cheat sheet on related rates and optimization. optimization related rates sample keywords maximize, minimize, […]

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]]> One of the main things students struggle with on exams is identifying the method to use on each problem. When they are learned in class, they are segregated by type, and their distinguishing features are not always highlighted. Here’s a brief cheat sheet on related rates and optimization.

optimization related rates
sample keywords maximize, minimize, most, smallest how fast, rate of change/increase/decrease
sample problem What dimensions maximize the area enclosed in a rectangular fence built with 100 feet of fencing? A shape is deforming such that it is always a rectangle of perimeter 100 ft, but the width is increasing by 5 ft/min. How fast is the area changing when the width is 20 feet?
goal maximize area find rate of change of area
answer form dimensions in feet rate in ft2/min
equation A = l\cdot w = (50-w)\cdot w = 50w - w^2 likewise, A = 50w - w^2
equation, unabridged A(w) = 50w - w^2 (area viewed as function of width) A(t) = 50w(t) - [w(t)]^2 (area and width viewed as related functions of time)
final answer a 25′ square fence maximizes the area the area is changing by 50 ft2/min

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]]> https://www.rweber.net/mathematics/calculus/distinguishing-related-rates-and-optimization/feed/ 2 272 Related Rates https://www.rweber.net/mathematics/calculus/related-rates/ https://www.rweber.net/mathematics/calculus/related-rates/#respond Mon, 25 Feb 2013 13:00:05 +0000 http://www.rweber.net/?p=258 These are perhaps the epitome of the dreaded “word problem.” Fortunately there is a step by step method for solving them. A. Assign variables to your quantities and find relationship equations for them. B. If relevant, simplify the equations to remove extraneous variables. C. Express given and asked-for rates of change in terms of d/dt. […]

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]]> These are perhaps the epitome of the dreaded “word problem.” Fortunately there is a step by step method for solving them.

A. Assign variables to your quantities and find relationship equations for them.
B. If relevant, simplify the equations to remove extraneous variables.
C. Express given and asked-for rates of change in terms of d/dt. At what point(s) do you want to know the rate(s) of change?
D. Differentiate relationship equation with respect to t.
E. Plug in known values and solve for desired value.

In C I will be finding the value of all variables at the desired point, although frequently you don’t need them.

Example 1.
The area of a square is increasing by 3 in2/s. What is the rate of change of the length of the sides when the side length is 2 in?
A. Name quantities: area A, side length \ell (same for all sides since it’s a square). Relationship: A = \ell^2.
B. This is already as simple as it gets.
C. Rates of change: given: \frac{dA}{dt} = 3 in2/s. We want to know \frac{d\ell}{dt} at the point \ell = 2, which is A = 4.
D. \frac{d}{dt}[A = \ell^2] is \frac{dA}{dt} = 2\ell \frac{d\ell}{dt}.
E. Know \frac{dA}{dt} = 3 and want \frac{d\ell}{dt} when \ell = 2; 3 = 4 \frac{d\ell}{dt} so \frac{d\ell}{dt} = \frac34. When the side length is 2 in, the sides are changing at a rate of 3/4 in/s.

Example 2.
You are testing walkie-talkie range with a friend. The two of you start at the same point, and then you amble north at 3 mph and your friend strides east at 4 mph. How fast are you moving away from each other when your friend is 3 mi away from the starting point?
A. Since you are moving north, label your position x, and since your friend is moving east, label her position y. You are really interested in the distance between you; call it R for (walkie-talkie) range. That is related to your positions by the equation R = \sqrt{x^2 + y^2}.
B. Since \frac{dx}{dt} and \frac{dy}{dt} are both constant, and hence x and y are constant multiples of each other, we could simplify by putting one in terms of the other but won’t gain much by doing so.
C. We are given \frac{dx}{dt} = 3 and \frac{dy}{dt} = 4 and want \frac{dR}{dt}. The point at which we want \frac{dR}{dt} is y=3, which is x = (\frac34)(3) = 2.25 and R = \frac{15}{4} = 3.75.
D. \frac{d}{dt}[R = \sqrt{x^2 + y^2}] is \frac{dR}{dt} = \frac12 (x^2+y^2)^{-1/2}(2x \frac{dx}{dt} + 2y \frac{dy}{dt}).
E. We know \frac{dx}{dt} = 3 and \frac{dy}{dt} = 4, and we want \frac{dR}{dt} when x = 2.25 and y = 3; \frac{dR}{dt} = (x^2+y^2)^{-1/2}(x \frac{dx}{dt} + y \frac{dy}{dt}) = (x \frac{dx}{dt} + y \frac{dy}{dt})/R = (2.25\cdot 3 + 3\cdot 4)/3.75 = 5. When you have gone 2.25 mi and your friend 3 mi, the distance between you is changing at a rate of 5 mph.

Example 3.
25 cubic inches of pizza dough are being spun in the air. Assume the surface is smooth and the outline is circular at all times. If the radius increases at a constant rate of 0.5 in/s, what is the rate of change of the thickness of the dough when the radius is 4 inches?
A. We have thickness, radius, and volume. Since t is taken, we’ll call them h, r, and V, respectively. The relationship is V = \pi r^2h.
B. Since V is constant, we can plug in its value. Since V is by itself on one side of the equation it is roughly equal amounts of work to do that as to leave it as V and plug in 0 for \frac{dV}{dt}, but since we did not simplify Example 2 we’ll simplify this one: 25 = \pi r^2 h.
C. We are given \frac{dr}{dt} = 0.5 in/s, and we want \frac{dh}{dt} at the point r = 4, which is h = \frac{25}{16\pi}.
D. \frac{d}{dt}[25 = \pi r^2h] is 0 = \pi (2r \frac{dr}{dt} h + r^2 \frac{dh}{dt}).
E. We know \frac{dr}{dt} = 0.5, and we want to know \frac{dh}{dt} when r = 4 and h = \frac{25}{16\pi}. Hence 0 = \pi (8 \cdot 0.5 \cdot \frac{25}{16\pi} + 16 \frac{dh}{dt}); -16\pi \frac{dh}{dt} = 100/16 = 25/4; \frac{dh}{dt} = -25/4 \cdot \frac{1}{16\pi} = -\frac{25}{64\pi} = -0.124 in/s. Make sense? If the volume is constant an increase in one dimension must be accompanied by a decrease in the other.

Example 4.
A coffee filter is being held under a running tap. It is cone-shaped, with height 8 cm and top radius 3 cm. Water flows in at the top at a rate of 4 cm3/s and drips out the bottom at 1.5 cm3/s. How fast is the water level in the filter rising when the depth is 2 cm?
A. Let’s call depth h since d is taken, radius r, and volume V. The volume equation for a cone is V = \frac{\pi}{3} r^2 h.
B. Because the filter is a cone, the ratio between depth and radius is the same for any amount of water in the filter. That is, while r is not always 3 and h is not always 8, r/h is always 3/8: r = 3/8 h. Thus the volume equation simplifies to V = \frac{3\pi}{64}h^3.
C. We are not given \frac{dV}{dt} right off, but the increase and decrease of water volume are both constant, so we can compute it: \frac{dV}{dt} = 4 - 1.5 = 2.5 cm3/s. We want to find \frac{dh}{dt} at the point h = 2, which gives V = \frac{3\pi}{8}.
D. \frac{d}{dt}[V = \frac{3\pi}{64}h^3] is \frac{dV}{dt} = \frac{9\pi}{64}h^2 \frac{dh}{dt}.
E. We have \frac{dV}{dt} = 2.5 and h = 2, so we obtain 2.5 = \frac{9\pi}{16} \frac{dh}{dt}, for \frac{dh}{dt} = \frac{40}{9\pi} = 1.41 cm/s. When the water is 2 cm deep, it is rising at a rate of 1.41 cm/s.

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