## mathcalculus 2 years ago hello guys, can someone help me with related rates? (attached)

1. Psymon

Eww, word problems *waits for the problem and prepares sword*

2. mathcalculus

3. Psymon

Oh, not a word problem :D okay, np with this one

4. Psymon

So whenever you take the derivative of y, you can mark it with (dy/dx). Whenever you take the derivative of an x, you can mark it with (d/dx). So take the derivative of both sides, put dy/dx and d/dx where necessary, and plug in the given values.

5. zzr0ck3r

the derivative of 7x^3 - 10x with respect to t Is 0

6. pgpilot326

dx/dt = 0 so will dy/dt

7. mathcalculus

alright let me try this step by step.

8. zzr0ck3r

y is in terms of x, there are no t's so treat x's like a constant derivative of a constant is 0

9. Psymon

Pretty much. They made it a little easy on ya with there being a dx/dt, haha. But itd be nice practice if we said it was d/dx and made d/dx equal to 1.

10. mathcalculus

okay so the derivative to this would be: 21x^2-10

11. Psymon

Correct.

12. zzr0ck3r

with respect to x, yes but they ask for it with respect to t (not x)

13. mathcalculus

okay. so do i multiply each by dx/dt?

14. Psymon

Its like theyre saying, there is no t for there to be a derivative in respect to, so everything is 0. Thats why I said we should change the conditions of the problem for practice.

15. zzr0ck3r

$\frac{d(f(x))}{dt}=0$

16. pgpilot326

x is some function of t, thus when we take the derivative of y with respect to t, we get $\frac{ dy }{ dt }=\frac{ d }{ dt }\left( 7x ^{3}-10x \right)=\frac{ d }{ dt }\left( 7x ^{3} \right)-\frac{ d }{ dt }\left( 10x \right)=21x ^{2}\frac{ dx }{ dt }-10\frac{ dx }{ dt }$form there it's plug in what you know and compute

17. pgpilot326

since dx/dt is 0, dy/dt is 0. y only depends on x. look at te function. if x doesn't change, neither will y. the math proves it.

18. mathcalculus

question: why would it be zero?

19. pgpilot326

the problem states that dx/dt = 0

20. pgpilot326

each term in dy/dt is multiplied by dx/dt

21. mathcalculus

and dy/dt= d/dt?

22. pgpilot326

no, d/dt is an operator. it says take the derivative of wht follows with respect to t.

23. pgpilot326

dy/dt is the derivative of y with respect to t.

24. pgpilot326

do you know the chain rule for derivatives?

25. mathcalculus

sort of, I need to refresh my memory with the chain rule.