Respuesta :
Answer:
[tex]\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\[/tex]
Hence, the slope , [tex]\dfrac{dy}{dx} = \dfrac{-1}{64}[/tex]
Step-by-step explanation:
We need to find the slope, i.e. [tex]\dfrac{dy}{dx}[/tex].
and all the functions are in terms of [tex]t[/tex].
So this looks like a job for the 'chain rule', we can write:
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)[/tex]
Given the functions
[tex]x = f(t)\\y = g(t)\\[/tex]
and
[tex]x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)[/tex]
we can differentiate them both w.r.t to [tex]t[/tex]
first we'll derivate Eq(B) to find dx/dt
[tex]x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\[/tex]
we can also rearrange Eq(B) to find x in terms of t , [tex]x = (37 - 4t^2)^{1/3}[/tex]. This is done so that [tex]\frac{dx}{dt}[/tex] is only in terms of t.
[tex]\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\[/tex]
we can find the value of this derivative using t = 3, and plug that value in Eq(A).
[tex]\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8[/tex]
now let's differentiate Eq(C) to find dy/dt
[tex]2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}[/tex]
rearrange Eq(C), to find y in terms of t, that is [tex]y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}[/tex]. This is done so that we can replace y in [tex]\frac{dy}{dt}[/tex] to make only in terms of t
[tex]\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\[/tex]
we can find the value of this derivative using t = 3, and plug that value in Eq(A).
[tex]\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}[/tex]
Finally we can plug all of our values in Eq(A)
but remember when plugging in the values that [tex]\frac{dy}{dt}[/tex] is being multiplied with [tex]\frac{dt}{dx}[/tex] and NOT [tex]\frac{dx}{dt}[/tex], so we have to use the reciprocal!
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}[/tex]
our slope is equal to [tex]\dfrac{-1}{64}[/tex]
