Parametric vs. Cartesian (vid#3): First Derivative dy/dx

TL;DR

This video explains how to find the derivative dy/dx and d^2y/dx^2 using parametric equations.

Transcript

okay in this video we're going to do some derivatives with parametric equations and before we start I just want to ask you that so subscribe if you guys haven't done it already and for my subscribers thank you so much okay so in particular we are focusing on getting dy/dx and E 2 ydx 2 when he is 4 if you refer back to the curve when T is 4 you hav... Read More

Key Insights

• 🐞 Differentiating the Cartesian equation for the top half of the graph helps find dy/dx.
• 👻 Parametric equations allow for a direct calculation of dy/dx and d^2y/dx^2.
• 🧘 Graphing the parametric equations helps understand the position and movement of the curve.

Questions & Answers

Q: How can we find dy/dx using the Cartesian equation for the top half of the graph?

By differentiating the equation Y = 2 + sqrt(x + 1), we can find dy/dx to be 1/(2sqrt(x + 1)). Substituting x=8 (when T=4) gives the value of 1/6.

Q: How do we find dy/dx using parametric equations?

We differentiate the parametric equation X = T^2 - 2T and find DX/DT = 2T - 2. Then, we differentiate the equation Y = T and find DY/DT = 1. Dividing DY/DT by DX/DT gives us an expression for dy/dx in terms of T.

Q: What does dy/dx represent geometrically?

dy/dx represents the slope of the tangent line to the curve at a specific point.

Q: What is d^2y/dx^2?

d^2y/dx^2 represents the second derivative of y with respect to x, which measures the rate of change of the slope of the tangent line.

Summary & Key Takeaways

• The video focuses on finding dy/dx and d^2y/dx^2 using parametric equations.

• Two methods are discussed: using the Cartesian equation for the top half of the graph and using the parametric equations directly.

• The process involves differentiation and substitution to find the values of dy/dx and d^2y/dx^2 at a specific point.