SOLUTION: find dy/dx x=t^2 y=square root of t^3 dx/dt= 2t dy/dt=square root of 3t^2 dy/dx= square root of 3t^2/2t (which is wrong, why?) x=t^2+5t y=t/5 dx/dt=2t+5 dy/dt=t5

Algebra ->  Functions -> SOLUTION: find dy/dx x=t^2 y=square root of t^3 dx/dt= 2t dy/dt=square root of 3t^2 dy/dx= square root of 3t^2/2t (which is wrong, why?) x=t^2+5t y=t/5 dx/dt=2t+5 dy/dt=t5      Log On


   

Question 885298: find dy/dx
x=t^2
y=square root of t^3
dx/dt= 2t
dy/dt=square root of 3t^2
dy/dx= square root of 3t^2/2t (which is wrong, why?)
x=t^2+5t
y=t/5
dx/dt=2t+5
dy/dt=t5^-1 =-5^-2t
dy/dx=(-5^-2)/(2t+5) which is of course, wrong again. please tutor, tell me what i am doing wrong!
thank you
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Are you allowed to eliminate t and then arrange into y=expressionOnlyInX, and then take derivative dy%2Fdx?

You showed two different pair of x & y equations.
I am working with the first pair!

Try this: dx%2Fdt=2t.

dy%2Fdt=%283%2F2%29t%5E%283%2F2-2%2F2%29=%283%2F2%29t%5E%281%2F2%29----(this is not appearing properly on the site)
The ratio of the y one to the x one could be dy%2Fdx.

dy%2Fdx=%28dy%2Fdt%29%2F%28dx%2Fdt%29

%28%283%2F2%29t%5E%281%2F2%29%29%2F%282t%29----(again this does not fully show on the site)

%283%2F4%29t%5E%281%2F2-1%29

%283%2F4%29t%5E%28-1%2F2%29

%283%2F%284t%5E%281%2F2%29%29%29
OR
highlight_green%28dy%2Fdx=%283%2F%284sqrt%28t%29%29%29%29