# Parametric Derivatives and Normal Equations for a Curve with Gradient 1

• sooyong94
In summary, the student found x and y and found that when y= -x- \frac{\pi}{4}, x+ y= \frac{\pi}{4}. They don't know how to determine if the normal at P meets the curve again.
sooyong94

## Homework Statement

The parametric equations of a curve are
##x=\frac{1}{2}(sint cost+ t), y=\frac{1}{2} t-\frac{1}{4} sin2t##,
##-\pi/2<t\leq0##. P is a point on the curve such that the gradient at P is 1. Find the equation of the normal at P. Hence, determine if the normal at P meets the curve again.

## Homework Equations

Parametric differentiation, chain rule

## The Attempt at a Solution

I have found ##\frac{dy}{dt}## and ##\frac{dt}{dx}##. ..Then ##\frac{dy}{dx}=\frac{1-cos 2t}{1+cos2t}##

I have also found out that when ##\frac{dy}{dx}=1##, ##t=\frac{-\pi}{4}##. Then the equation of the normal is ##y=-x-\frac{\pi}{4}##. Now I don't know how to determine if the normal at P meets the curve again...

hi sooyong94!
sooyong94 said:
I have also found out that when ##\frac{dy}{dx}=1##, ##t=\frac{-\pi}{4}##. Then the equation of the normal is ##y=-x-\frac{\pi}{4}##. Now I don't know how to determine if the normal at P meets the curve again...

hint: what is x + y ?

t ? :P

Yes, using the fact that sin(2t)= 2sin(t)cos(t), adding the parametric equations for x and y gives x+ y= t.
But $y= -x- \frac{\pi}{4}$ tells you that $x+ y= \frac{\pi}{4}$. Wherever the curve and line intersect, they must both be true.

HallsofIvy said:
Yes, using the fact that sin(2t)= 2sin(t)cos(t), adding the parametric equations for x and y gives x+ y= t.
But ##y= -x- \frac{\pi}{4}## tells you that ##x+ y= \frac{\pi}{4}##. Wherever the curve and line intersect, they must both be true.

But that's ##t=\frac{\pi){4}##, which is not within the domain of t. :/

sooyong94 said:
But that's ##t=\frac{\pi){4}##,

no, its -π/4

:hmm: Wasn't there should another coordinate that intersects with the curve other than x=-pi/4?

??

do you mean x+y = -π/4 ?​

I mean the normal line cuts the curve at other points... is it possible for that?

Questions about derivatives belong in the Calculus & Beyond section...

## 1. What is a parametric derivative?

A parametric derivative is a mathematical concept used to find the rate of change of a variable in a parametric equation. It is similar to a regular derivative, but it involves differentiating both the x and y variables with respect to a third variable, typically denoted as t.

## 2. How is a parametric derivative calculated?

To calculate a parametric derivative, you must first differentiate both the x and y variables with respect to t. Then, use the chain rule to multiply the derivatives with respect to t by dt/dx or dt/dy, depending on which variable you are solving for. Finally, divide the two derivatives to get the parametric derivative.

## 3. Why are parametric derivatives important?

Parametric derivatives are important in many fields of science and engineering because they allow us to calculate the rate of change of a variable in a parametric equation. This is useful for understanding and predicting the behavior of systems that involve changing quantities, such as motion, growth, and decay.

## 4. Can parametric derivatives be negative?

Yes, parametric derivatives can be negative. A negative value indicates that the variable is decreasing with respect to the third variable, while a positive value indicates that the variable is increasing. The magnitude of the derivative represents the rate of change, so a larger negative value indicates a faster decrease, while a smaller negative value indicates a slower decrease.

## 5. How are parametric derivatives used in real-world applications?

Parametric derivatives are commonly used in physics, engineering, and economics to model and analyze changing systems. For example, in physics, parametric derivatives are used to calculate the velocity and acceleration of an object in motion, while in economics, they can be used to model the growth or decline of a company's revenue over time.

• Calculus and Beyond Homework Help
Replies
3
Views
891
• Calculus and Beyond Homework Help
Replies
20
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
221
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
711
• Calculus and Beyond Homework Help
Replies
1
Views
916
• Calculus and Beyond Homework Help
Replies
5
Views
669
• Calculus and Beyond Homework Help
Replies
2
Views
987
• Calculus and Beyond Homework Help
Replies
13
Views
453