Product of a Derivative and its Inverse

In summary, the conversation is about calculating the derivative of θ with respect to x and simplifying it. It is believed that the result should always be 1, but the process of solving it leads to a messy expression. The solution is found by substituting x for r*cos(θ) and simplifying the resulting expression to sin(θ)/√(1-x^2/r^2). The conversation also emphasizes the importance of using parentheses when writing fractional expressions in one line.
  • #1
EconStudent
4
0

Homework Statement



For the functions:

x = r*cos(θ)
y = r*sin(θ)

Calculate dθ/dX * dX/dθ (considering θ as a function of x, y) and simplify.

The Attempt at a Solution



I believe this should always be 1, by definition (as the product of a derivative and its inverse). However, I don't know how to show this.

Solving the first function for θ, I get cos-1(x/r). The derivative of this is a relatively messy expression 1 over the product of r and a square root, and its being multiplied by something relatively simple (-r*sin(θ)). Why is it that this would simplify to 1?

I get that the negatives and the r's will cancel. So it simplifies to:

sin(θ) / √(1 - x2/r2)

But I can't get any further nor am I satisfied that this expression is or is not equal to 1. Thanks in advance.
 
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  • #2
EconStudent said:

Homework Statement



For the functions:

x = r*cos(θ)
y = r*sin(θ)

Calculate dθ/dX * dX/dθ (considering θ as a function of x, y) and simplify.

The Attempt at a Solution



I believe this should always be 1, by definition (as the product of a derivative and its inverse). However, I don't know how to show this.

Solving the first function for θ, I get cos-1(x/r). The derivative of this is a relatively messy expression 1 over the product of r and a square root, and its being multiplied by something relatively simple (-r*sin(θ)). Why is it that this would simplify to 1?

I get that the negatives and the r's will cancel. So it simplifies to:

sin(θ) / √(1 - x2/r2)

But I can't get any further nor am I satisfied that this expression is or is not equal to 1. Thanks in advance.

Use that x/r=cos(theta). What does that make 1-(x/r)^2?
 
  • #3
That's so clever, thank you. Substituting for x and squaring the top and bottom, we get sin2/1-cos2 which is just sin2/sin2.
 
  • #4
EconStudent said:
That's so clever, thank you. Substituting for x and squaring the top and bottom, we get sin2/1-cos2 which is just sin2/sin2.
When you write fractional expressions in a line, and there are multiple terms in the top or bottom, USE PARENTHESES!

What you wrote would be reasonably interpreted as
[tex]\frac{sin^2(x)}{1} -cos^2(x)[/tex]
 

What is the product of a derivative and its inverse?

The product of a derivative and its inverse refers to the result obtained when multiplying a derivative function and its inverse function. This product is equivalent to the identity function, meaning that it will return the original input value.

Why is the product of a derivative and its inverse important?

The product of a derivative and its inverse plays a crucial role in calculus and mathematical analysis. It is used to solve various problems, such as finding the slope of a tangent line, calculating the area under a curve, and determining the rate of change of a function.

How is the product of a derivative and its inverse calculated?

The product of a derivative and its inverse can be calculated using the chain rule in calculus. This involves taking the derivative of the original function, then multiplying it by the derivative of its inverse function.

Can the product of a derivative and its inverse ever be equal to a constant?

No, the product of a derivative and its inverse will never be equal to a constant. This is because the derivative of a function is a measure of its rate of change, and the inverse function undoes the effect of the original function. Therefore, the product will always result in the original input value.

What are some real-life applications of the product of a derivative and its inverse?

The product of a derivative and its inverse has various applications in fields such as physics, engineering, and economics. For example, it can be used to determine the maximum profit or minimum cost in economics, the velocity of a moving object in physics, and the maximum power output of an electrical circuit in engineering.

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