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

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) (Show Source): You can put this solution on YOUR website!
Are you allowed to eliminate t and then arrange into , and then take derivative ?

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

Try this: .

----(this is not appearing properly on the site)
The ratio of the y one to the x one could be .

----(again this does not fully show on the site)

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