Derivative with a range constraint - mystified

This is not the first time I have seen a calculus or differential equations problem that was misstated. Some well known calculus texts are lousy examples of clear exposition and many totally fail at giving accurate and sufficient definitions of terms.In summary, the problem is to show that da/dx = cos^3(b)cos(a) / cos(a+b)cos(a-b). The attempt at a solution involves manipulating the equation to get rid of the sin^2(a)sin^2(b) term in the denominator, but there seems to be a counterexample that disproves the equation. This suggests that the question may be wrong or that there may be some additional consideration needed to take into account the range of a, b, and x.
  • #1
Gekko
71
0

Homework Statement



x=sin(a) / cos(b)

a+b < pi/2
a>0, b>0
0<x<1

show that

da/dx = cos^3(b)cos(a) / cos(a+b)cos(a-b)


The Attempt at a Solution



dx/da = cos(a) / cos(b) therefore
da/dx = cos(b) / cos(a)

=cos^3(b)cos(a) / cos^2(b)cos^2(a)

However the denominator of the desired format = cos(a+b)cos(a-b) = cos^2(a)cos^2(b)-sin^2(a)sin^2(b)

Not sure how to get rid of the sin^2(a)sin^2(b) term.

Is the question wrong or is there something special that needs to be done to take into account the range of a, b and x?

Very much appreciate help as I've been totally stumped on this and the equality is required further on
 
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  • #2
If you add and subtract cos2(a)sin2(b) in the denominator your equality comes out:

[tex]\frac{\cos(b)}{\cos(a)}=\frac{\cos^3(b)\cos(a)}{\cos^2(a)-\sin^2(b)}[/tex]

At least I think it does; you can check it. The problem is if you let a = b = π/6, you get the left side = 0.9742785794, the right side = 1, and x = 0.5773502695.

So it would appear something is amiss. :frown:
 
  • #3
This seems to be the same question you posted in this thread.
This equation is not an identity
[tex]\frac{cos(b)}{cos(a)}=\frac{cos^3(b)cos(a)}{cos^2(a)-sin^2(b)}[/tex]
LCKurtz points out a counterexample. Another is a = π/6 and b = π/4. Using these numbers the left side value is sqrt(6)/3 and the right side value is sqrt(6)/2.
 

FAQ: Derivative with a range constraint - mystified

What is a derivative with a range constraint?

A derivative with a range constraint is a mathematical concept that involves finding the rate of change of a function at a specific point, while also taking into account a given range of values that the function must stay within.

Why is a range constraint important in taking a derivative?

A range constraint is important in taking a derivative because it allows us to consider the context in which the function is being used. This can be useful in real-world applications where certain limits or restrictions are present.

How do you solve for a derivative with a range constraint?

To solve for a derivative with a range constraint, you must first find the derivative of the function using standard rules and techniques. Then, you must check if the resulting derivative satisfies the given range constraint. If it does not, you may need to adjust the original function to meet the constraint, and then recalculate the derivative.

What are some examples of functions with range constraints?

Functions with range constraints can include physical quantities such as velocity or acceleration, where the values must fall within a certain range. They can also include economic models where certain variables must remain within a specific range. For example, a company's production cannot exceed its maximum capacity.

How is a derivative with a range constraint useful in real life?

A derivative with a range constraint can be useful in real life situations where we need to consider not only the rate of change of a function, but also the limitations or restrictions that may be present. This can help us make more accurate predictions or decisions, such as in financial planning or engineering design.

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