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mathcalculus
hello guys, can someone help me with related rates? (attached)
Eww, word problems *waits for the problem and prepares sword*
Oh, not a word problem :D okay, np with this one
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.
the derivative of 7x^3 - 10x with respect to t Is 0
dx/dt = 0 so will dy/dt
alright let me try this step by step.
y is in terms of x, there are no t's so treat x's like a constant derivative of a constant is 0
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.
okay so the derivative to this would be: 21x^2-10
with respect to x, yes but they ask for it with respect to t (not x)
okay. so do i multiply each by dx/dt?
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.
\[\frac{d(f(x))}{dt}=0\]
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
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.
question: why would it be zero?
the problem states that dx/dt = 0
each term in dy/dt is multiplied by dx/dt
no, d/dt is an operator. it says take the derivative of wht follows with respect to t.
dy/dt is the derivative of y with respect to t.
do you know the chain rule for derivatives?
sort of, I need to refresh my memory with the chain rule.