In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using y = f ( x ) to represent a function, an applied problem might use C = C ( q ) to represent the cost C of manufacturing q units of a good. Because of this, the inverse notation f − 1 used in a pure mathematics problem is not used when finding inverses of applied problems. Rather, the inverse of a function such as C = C ( q ) will be q = q ( C ) . So C = C ( q ) is a function that represents the cost C as a function of the number q of units manufactured, and q = q ( C ) is a function that represents the number q as a function of the cost C . Problems 91-94 illustrate this idea. Ideal Body Weight One model for the ideal body weight W for men (in kilograms) as a function of height h (in inches) is given by the function W ( h ) = 50 + 2.3 ( h − 60 ) (a) What is the ideal weight of a 6-foot male? (b) Express the height h as a function of weight W . (c) Verify that h = h ( W ) is the inverse of W = W ( h ) by showing that h ( W ( h ) ) = h and W ( h ( W ) ) = W . (d) What is the height of a male who is at his ideal weight of 80 kilograms? [ Note: The ideal body weight W for women (in kilograms) as a function of height h (in inches) is given by W ( h ) = 45.5 + 2.3 ( h − 60 ) .
In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using y = f ( x ) to represent a function, an applied problem might use C = C ( q ) to represent the cost C of manufacturing q units of a good. Because of this, the inverse notation f − 1 used in a pure mathematics problem is not used when finding inverses of applied problems. Rather, the inverse of a function such as C = C ( q ) will be q = q ( C ) . So C = C ( q ) is a function that represents the cost C as a function of the number q of units manufactured, and q = q ( C ) is a function that represents the number q as a function of the cost C . Problems 91-94 illustrate this idea. Ideal Body Weight One model for the ideal body weight W for men (in kilograms) as a function of height h (in inches) is given by the function W ( h ) = 50 + 2.3 ( h − 60 ) (a) What is the ideal weight of a 6-foot male? (b) Express the height h as a function of weight W . (c) Verify that h = h ( W ) is the inverse of W = W ( h ) by showing that h ( W ( h ) ) = h and W ( h ( W ) ) = W . (d) What is the height of a male who is at his ideal weight of 80 kilograms? [ Note: The ideal body weight W for women (in kilograms) as a function of height h (in inches) is given by W ( h ) = 45.5 + 2.3 ( h − 60 ) .
Solution Summary: The author explains the ideal body weight W for men as a function of height h (in inches).
In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using
to represent a function, an applied problem might use
to represent the cost
of manufacturing q units of a good. Because of this, the inverse notation
used in a pure mathematics problem is not used when finding inverses of applied problems. Rather, the inverse of a function such as
will be
. So
is a function that represents the cost
as a function of the number
of units manufactured, and
is a function that represents the number
as a function of the cost
. Problems 91-94 illustrate this idea.
Ideal Body Weight One model for the ideal body weight
for men (in kilograms) as a function of height
(in inches) is given by the function
(a) What is the ideal weight of a 6-foot male?
(b) Express the height
as a function of weight
.
(c) Verify that
is the inverse of
by showing that
and
.
(d) What is the height of a male who is at his ideal weight of 80 kilograms?
[Note: The ideal body weight
for women (in kilograms) as a function of height
(in inches) is given by
.
2. Answer the following questions.
(A) [50%] Given the vector field F(x, y, z) = (x²y, e", yz²), verify the differential identity
Vx (VF) V(V •F) - V²F
(B) [50%] Remark. You are confined to use the differential identities.
Let u and v be scalar fields, and F be a vector field given by
F = (Vu) x (Vv)
(i) Show that F is solenoidal (or incompressible).
(ii) Show that
G =
(uvv – vVu)
is a vector potential for F.
A driver is traveling along a straight road when a buffalo runs into the street. This driver has a reaction time of 0.75 seconds. When the driver sees the buffalo he is traveling at 44 ft/s, his car can decelerate at 2 ft/s^2 when the brakes are applied. What is the stopping distance between when the driver first saw the buffalo, to when the car stops.
Topic 2
Evaluate S
x
dx, using u-substitution. Then find the integral using
1-x2
trigonometric substitution. Discuss the results!
Topic 3
Explain what an elementary anti-derivative is. Then consider the following
ex
integrals: fed dx
x
1
Sdx
In x
Joseph Liouville proved that the first integral does not have an elementary anti-
derivative Use this fact to prove that the second integral does not have an
elementary anti-derivative. (hint: use an appropriate u-substitution!)
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY