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Tresset [83]
3 years ago
9

Assuming that the equation defines x and y implicitly as differentiable functions xequals​f(t), yequals​g(t), find the slope of

the curve xequals​f(t), yequals​g(t) at the given value of t. x cubed plus 4 t squaredequals37​, 2 y cubed minus 2 t squaredequals110​, tequals3
Mathematics
1 answer:
Doss [256]3 years ago
7 0

Answer:

\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\

Hence, the slope , \dfrac{dy}{dx} = \dfrac{-1}{64}

Step-by-step explanation:

We need to find the slope, i.e. \dfrac{dy}{dx}.

and all the functions are in terms of t.

So this looks like a job for the 'chain rule', we can write:

\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)

Given the functions

x = f(t)\\y = g(t)\\

and

x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)

we can differentiate them both w.r.t to t

first we'll derivate Eq(B) to find dx/dt

x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\

we can also rearrange Eq(B) to find x in terms of t , x = (37 - 4t^2)^{1/3}. This is done so that \frac{dx}{dt} is only in terms of t.

\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

\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

now let's differentiate Eq(C) to find dy/dt

2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}

rearrange Eq(C), to find y in terms of t, that is y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}. This is done so that we can replace y in \frac{dy}{dt} to make only in terms of t

\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that \frac{dy}{dt} is being multiplied with \frac{dt}{dx} and NOT \frac{dx}{dt}, so we have to use the reciprocal!

\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}

our slope is equal to \dfrac{-1}{64}

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