Given:
x and y are both differentiable functions of t.
[tex]4x^2+3y^3=28[/tex]
[tex]x=-1\text{ and }\dfrac{dy}{dt}=8[/tex]
To find:
The value of [tex]\dfrac{dx}{dt}[/tex].
Solution:
We have,
[tex]4x^2+3y^3=28[/tex] ...(i)
At x=-1,
[tex]4(-1)^2+3y^3=28[/tex]
[tex]4+3y^3=28[/tex]
[tex]3y^3=28-4[/tex]
[tex]3y^3=24[/tex]
Divide both sides by 3.
[tex]y^3=8[/tex]
Taking cube root on both sides.
[tex]y=2[/tex]
So, y=2 at x=-1.
Differentiate (i) with respect to t.
[tex]4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0[/tex]
Putting x=-1, y=2 and [tex]\dfrac{dy}{dt}=8[/tex], we get
[tex]4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0[/tex]
[tex]-8\dfrac{dx}{dt}+9(4)(8)=0[/tex]
[tex]-8(\dfrac{dx}{dt}-9(4))=0[/tex]
Divide both sides by -8.
[tex]\dfrac{dx}{dt}-36=0[/tex]
[tex]\dfrac{dx}{dt}=36[/tex]
Therefore, the value of [tex]4x^2+3y^3=28[/tex] is 36.
