Derivatives of Parametric Functions

This calculus 2 video tutorial explains how to find the derivative of a parametric function.
Parametric Equations - Introduction: • Parametric Equations I...
Derivatives of Parametric Functions:
• Derivatives of Paramet...
Tangent Lines of Parametric Curves:
• Tangent Lines of Param...
2nd Derivative - Parametric Equations:
• Second Derivatives of ...
3rd Derivative of Parametric Curves:
• Differentiation of Par...
___________________________________
Horizontal & Vertical Tangent Lines:
• Horizontal Tangent Lin...
Area of Parametric Curves:
• Area of Parametric Curves
Arc Length of Parametric Curves:
• Arc Length of Parametr...
Surface Area of Parametric Curves:
• Surface Area of Revolu...
Polar Coordinates - Introduction:
• Polar Coordinates Basi...
Tangent Line Equations - Polar Form:
• Tangent Line Equations...
____________________________________
Horizontal & Vertical Tangents - Polar:
• Horizontal Tangent Lin...
Area of Polar Curves:
• Finding Area In Polar ...
Area Between Polar Curves:
• Finding Area Bounded B...
Arc Length of Polar Curves:
• Arc Length of Polar Cu...
Surface Area of Polar Curves:
• Surface Area of Revolu...
_______________________________________
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[@TheOrganicChemistryTutor]

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Honestly I tried searching for a video that I could understand for parametric equations but I couldn't for the life of me! And then I come across this and I'm able to solve everything in a matter of minutes! Thank you so much

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Great video, you really helped solidify my intuition about parametric equations (which are the best kind imo!)

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[@alastairhewitt380]

Thank you once again for quality content, but it would have been really helpful if you showed us how to take the second derivative as well.

[@missubappapudchyawarshiluk1310]

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[@bartimaeu1]

For second derivative don't you just find the derivative of the derivative? And since it's usually a simple, standard equation at that point it should be easy

[@user-cl5sn1gg6i]

Can we obtain back the two original parametric functions by finding the antiderivative of dy/dx?

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[@richard5814]

Is the answer of 4-15t also correct in example 2? 4:10

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[@tonito5588]

I'm kinda confused so this might be a dumb question; but why can't you just rewrite y in terms of x, then take the derivative of that function? Is it just easier to take the derivative in terms of t?

[@tomatrix7525]

If I understand your Question properly, you actually could, for example look at the question at 3:20 where we must get dy/dx. Going by your method, we could rewrite tye function y=6t^2 -15t^3 in terms of x. To do this we must express t in terms of x. X =t^3 as given. So thereforr we could simply get dy/dx when we rewrite y(x) as a function of x. If x=t^3 then t = cube root of x Now for y= 6t^2 ......etc replace t for cb. Rt of x Derrive this new function and this IS dy/dx ( meaning derivative of function y in terms of x) This is an equivalently correct method and would be correct too!

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[@kdixon-campbell1774]

Always liked your videos, but the parametric equations y=sec and x=18tan -5, AFTER the differentiation, I do not agree with dividing sec² with 1 of the sec in the denominator that was being added with tan....

[@estevesamielandrew4257]

is it always letter "t"? Or it depends on the variable?

[@shriyansh.]

bruh.......................................... depends on the variable

[@lawrencemills1355]

At 7:42 is 1/sin not cosec, you said it is secant and I got a bit confused

[@kathyk7484]

how does one know the derivatives?

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[@gamedroidtech9866]

Hey, sir can't we cut T by T here at 5:08??

[@floatingferry6238]

No u can't, because 't' represents angle value of sin and also cos.

[@sandrasarpong7148]

I don’t understand why you didn’t cancel out the T please

[@AaaBbb-kw5em]

@@sandrasarpong7148 it's like saying sin(30)/cos(30) can you cancel/cut both the (30)'s here? No you can't. You can however call it tan(30) which is the same as what happened in the video here.

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