Question

Can someone tell me why, when you take the derivative of a parametric equation, it's the derivative wrt t of dy divided by the derivative wrt t of dx

Answer 1

\[\Large \frac{ dy }{ dt } \div \frac{ dx }{ dt }=\frac{ dy }{ dt } \times \frac{ dt }{ dx }\approx \frac{ dy }{ dx }\]

Answer 2

chain rule

Answer 3

I understand that much, and I put approximation symbols (even though it's actually an equal sign), but thats the part that gets me troubled. It's the same for chain rule, I never really understood the theory behind it and every chain rule proof flies over my head. Why is it though, that this derivative is able to be split into two?

Answer 4

It's more of a chain rule question, yes I admit @DanJS

Answer 5

The real question is Are those dts actually even there? Or are there actually a bunch of dts, dps, drs, etc.. hidden in the derivative for how many functions you have, if it's only one function like x then there isn't any dts etc?

Answer 6

\[\text{ let } y=f(x) \\ \frac{dy}{dx}=f'(x) \\ dy=f'(x) dx \\ \frac{dy}{dt} \cdot \frac{dt}{dx} \\ =\frac{f'(x) dx}{dt} \cdot \frac{dt}{dx} =\frac{f'(x) \cancel{dx} }{dt} \cdot \frac{dt}{\cancel{dx}} =f'(x) \frac{dt}{dt}=f'(x) \cdot 1=f'(x)=\frac{dy}{dx}\]

Answer 7

sorry,

Answer 8

yeah, that looks good, the more important thing is to understand parametric form, the derivatives are just like freckles wrote

Answer 9

It is strange multiplying and dividing by differentials at first...

Answer 10

u in calc 3?

Answer 11

vectors
r(t) = x(t) + y(t)

Answer 12

Calc BC, this only counts for half a semester of college math, so I think calc II

Answer 13

\[\text{ I don't know if you need an example of this but I will give one just \in case } \\ \text{ say we have the parametric curves } x=t^2 \text{ and } y=t-1 \\ \frac{dx}{dt}=2t \text{ and } \frac{ dy}{dt}=1 \\ \text{ so } \frac{dt}{dx}=\frac{1}{2t} \\ \\ \frac{dy}{dx}=\frac{dt}{dx} \cdot \frac{dy}{dt}=\frac{1}{2t} \cdot 1 =\frac{1}{2t} \]
we can get this same derivative getting rid of the parameter t and then differentiating
\[x=(y+1)^2 \\ 1=2(y+1)^{2-1}(\frac{dy}{dx}+0) \\ 1=2(y+1)^1(\frac{dy}{dx}) \\ 1=2(y+1)\frac{dy}{dx} \\ \frac{1}{2(y+1)} =\frac{dy}{dx} \\ \text{ recall } y+1=t \\ \frac{1}{2t}=\frac{dy}{dx}\]

Answer 14

Yeah probably the only thing you have to remember is the chain rule thing mentioned above

Answer 15

I understand how to do them, just theory etc has me confused.

Answer 16

x and Y are functions of a parameter t, so to find the change in Y per the change in X, you need to find dy/dt and dt/dx, and dx/dt and dt/dx

Answer 17

could combine them like above and use implicit differentiation instead

Answer 18

which is still the chain rule

Answer 19

I think the most important part is to understand parametric equations, how both the y ad x coordinates are determined by a different variable.. like t time for example

Answer 20

so to get how y changes with x, you need to understand how both y and x change with time t

Answer 21

like a 2 dimensional position w.r.t. time

Answer 22

Usually use vectors r(t) = x(t) + y(t), for a position

Answer 23

Brent has two barcode printing machines. Each machine can print 25 barcodes per minute. On day 1, only the first machine was used, and it printed 95 barcodes. On the following day, the first machine printed for x minutes and the second machine printed for y minutes. Which expression shows the total number of barcodes printed by both machines during these two days?

Answer 24

please help me on this