132k views
2 votes
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

1 Answer

1 vote

Answer:


(dx)/(dt) = -8,(dy)/(dt) = 1/8\\

Hence, the slope ,
(dy)/(dx) = (-1)/(64)

Explanation:

We need to find the slope, i.e.
(dy)/(dx).

and all the functions are in terms of
t.

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


(dy)/(dx) = (dy)/(dt) .(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(dx)/(dt) + 8t = 0\\(dx)/(dt) = (-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
(dx)/(dt) is only in terms of t.


(dx)/(dt) = (-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).


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

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


2y^3 - 2t^2 = 110\\6y^2(dy)/(dt) -4t = 0\\(dy)/(dt) = (4t)/(6y^2)

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


(dy)/(dt) = (4t)/(6y^2)\\(dy)/(dt)=(4t)/(6\left((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).


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

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

but remember when plugging in the values that
(dy)/(dt) is being multiplied with
(dt)/(dx) and NOT
(dx)/(dt), so we have to use the reciprocal!


(dy)/(dx) = (dy)/(dt) .(dt)/(dx)\\(dy)/(dx) = (1)/(8).(1)/(-8) \\(dy)/(dx) = (-1)/(64)

our slope is equal to
(-1)/(64)

User Ott Toomet
by
5.6k points