Answer:
![(dx)/(dt) = -8,(dy)/(dt) = 1/8\\](https://img.qammunity.org/2020/formulas/mathematics/college/dzgp7j3gcnu2r8n0any2tk38c9v4jo9dnm.png)
Hence, the slope ,
![(dy)/(dx) = (-1)/(64)](https://img.qammunity.org/2020/formulas/mathematics/college/p0sp3jwlj2tr48hc394gsopz0njzv3cmr5.png)
Explanation:
We need to find the slope, i.e.
.
and all the functions are in terms of
.
So this looks like a job for the 'chain rule', we can write:
Given the functions
![x = f(t)\\y = g(t)\\](https://img.qammunity.org/2020/formulas/mathematics/college/v8sjdskymegl0h604s15y982uvr8jtodw3.png)
and
![x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)](https://img.qammunity.org/2020/formulas/mathematics/college/ki9tlylo4y8vkrl6azw9oge24hfwn0vdj5.png)
we can differentiate them both w.r.t to
![t](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wv4bwbdwodymhjzkyfun99mhtm6mryycmu.png)
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)\\](https://img.qammunity.org/2020/formulas/mathematics/college/l34voic8llhf83r20y85admtf92hdejr9g.png)
we can also rearrange Eq(B) to find x in terms of t ,
. This is done so that
is only in terms of t.
![(dx)/(dt) = (-8t)/(3(37 - 4t^2)^(2/3))\\](https://img.qammunity.org/2020/formulas/mathematics/college/u21vj45vr4lvglw8exnj7dmz68y5nsdkwv.png)
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](https://img.qammunity.org/2020/formulas/mathematics/college/zbd58pz381ldiv33nsdvi7l85r738xc2b6.png)
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)](https://img.qammunity.org/2020/formulas/mathematics/college/gekljhid794ezthlvrkmu5q2m3a5cl5p27.png)
rearrange Eq(C), to find y in terms of t, that is
. This is done so that we can replace y in
to make only in terms of t
![(dy)/(dt) = (4t)/(6y^2)\\(dy)/(dt)=(4t)/(6\left((110 + 2t^2)/(2)\right)^(2/3))\\](https://img.qammunity.org/2020/formulas/mathematics/college/z9z8z8ojaqac7nl9sug120gip6tukk3r67.png)
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)](https://img.qammunity.org/2020/formulas/mathematics/college/pc56sc1wqu7thvc2ljl87fuu3ky9i1men5.png)
Finally we can plug all of our values in Eq(A)
but remember when plugging in the values that
is being multiplied with
and NOT
, so we have to use the reciprocal!
![(dy)/(dx) = (dy)/(dt) .(dt)/(dx)\\(dy)/(dx) = (1)/(8).(1)/(-8) \\(dy)/(dx) = (-1)/(64)](https://img.qammunity.org/2020/formulas/mathematics/college/2e5upi5kdvnow5ydj8qxxv5f060ashjo2s.png)
our slope is equal to
![(-1)/(64)](https://img.qammunity.org/2020/formulas/mathematics/college/l3lwiz9p3dg2tz0gff98lwgv7x2vsiqjpg.png)