Derivative of parametric function

In summary, the conversation discusses finding the tangent line and second derivative at a given point on a curve. The first derivative is found to be -sqrt3 sint/-sint, which simplifies to sqrt3 and matches the calculator's result. The equation of the tangent line is determined to be y=sqrt3 x. The speaker is having trouble with the second derivative and believes it to be 0. Another person confirms this and provides an alternative method of finding the second derivative.
  • #1
mastiffcacher
25
0

Homework Statement


Find the line tangent to the point 2pi/3 when x=cost y=sqrt3 cost.
Also find the value of d2y/dx2 at the point given.


Homework Equations


I found dy/dx to be -sqrt3 sint/ -sint. I found that to be just sqrt3. This matched what my calculator told me the slope was. The equation of the tangent is y=sqrt3 x or at least I think.

I am having trouble with the second derivative. I believe that it is 0. When I applied the quotient rule to dy/dx I got sqrt3 sintcost-sqrt3 sintcost/sin2t. I then multiplied by dt/dx. Either way, the numerator is still 0.

Am I right, on the right path, or completely wrong.
 
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  • #2
I think you're right. d2y/dx2 = 0
 
  • #3
Yes that is correct. You can get the same answer easily by noting that since dy/dx = constant. Hence d/dx (dy/dx) = 0.
 

Related to Derivative of parametric function

1. What is a parametric function?

A parametric function is a mathematical expression that defines the relationship between two or more variables. Unlike a regular function, a parametric function has its input variables (usually denoted by t) depend on a third parameter, often time.

2. How do you take the derivative of a parametric function?

To take the derivative of a parametric function, you need to use the chain rule. First, differentiate each individual variable with respect to the parameter t. Then, multiply each derivative by dt and add them together to get the total derivative of the parametric function.

3. What is the purpose of finding the derivative of a parametric function?

The derivative of a parametric function can represent the rate of change of the function with respect to the parameter t. This can be useful in determining the slope of a curve at a specific point, the velocity of an object, or the acceleration of a moving body.

4. Can you take the derivative of both the x and y components of a parametric function?

Yes, you can take the derivative of both the x and y components of a parametric function. This is known as a vector derivative and is commonly used in physics and engineering to calculate quantities such as velocity and acceleration.

5. Are there any special cases when taking the derivative of a parametric function?

Yes, there are special cases when taking the derivative of a parametric function. One example is when the parametric function is given in terms of polar coordinates. In this case, the derivative involves using the product rule and converting the polar coordinates to Cartesian coordinates.

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