Let z a function of x,y near (2,1,-1) that is determined by $\displaystyle x^2z+xyz^3=-6$. If $\displaystyle x(t)=t^3-3t$ and $\displaystyle y(t) =3t^2-6t+1$, find $\displaystyle \frac{dz}{dt}$ for $\displaystyle t=2$.
Let z a function of x,y near (2,1,-1) that is determined by $\displaystyle x^2z+xyz^3=-6$. If $\displaystyle x(t)=t^3-3t$ and $\displaystyle y(t) =3t^2-6t+1$, find $\displaystyle \frac{dz}{dt}$ for $\displaystyle t=2$.
Thank you for your response. Could you please check the following?
1. Substitute the functions of t into z.
$\displaystyle (t^3-3t)^2z+(t^3-3t)(3t^2-6t+1)z^3=-6$
2. Differentiate both sides implicitly with respect to t.
$\displaystyle 2(t^3-3t)(3t^2-3)z + (t^3-3t)^2 \frac{dz}{dt} + \frac{d[(3t^5-6t^4+t^3-9t^3+18t^2-3t)z^3 ]}{dt}=0$
For t=2:
$\displaystyle 2(2^3-3(2))(3(2)^2-3)z + (2^3-3(2))^2 \frac{dz}{dt} +\frac{d[(3(2)^5-6(2)^4+2^3-9(2)^3+18(2)^2-3(2)) z^3]}{dt}=0\Rightarrow2z(2)(9)+4\frac{dz}{dt}+\frac{d[(96-96+8-72+72-6) z^3 ]}{dt}=0\Rightarrow36z+4\frac{dz}{dt}+\frac{d[2z^3]}{dt}=0$
3. Solve for $\displaystyle \frac{dz}{dt}$.
$\displaystyle 36z + 4 \frac{dz}{dt} + 6z^2\frac{dz}{dt}=0\Rightarrow\frac{dz}{dt}=\frac{-36z}{6z^2+4}$
For z=-1:
$\displaystyle \frac{dz}{dt} (z=-1) = \frac{-36(-1)}{6(-1)^2+4}=\frac{36}{10}=3.6$
Yes, I should have evaluated the derivative first! The right solution is the following:
$\displaystyle 2(t^3-3t)(3t^2-3)z+(t^3-3t)^2\frac{dz}{dt}+(15t^4-24t^3+3t^2-27t^2+36t-3)z^3+(3t^5-6t^4+t^3-9t^3+18t^2-3t)3z^2\frac{dz}{dt}=0$
Now substituting t=2 gives:
$\displaystyle 2(2^3-3(2))(3(2)^2-3)z+(2^3-3(2))^2\frac{dz}{dt}+(15(2)^4-24(2)^3+3(2)^2-27(2)^2+36(2)-3)z^3+(3(2)^5-6(2)^4+(2)^3-9(2)^3+18(2)^2-3(2))3(-1)^2\frac{dz}{dt}$
$\displaystyle =>36z+4\frac{dz}{dt}+21z^3+6z^2\frac{dz}{dt}$
Solving for $\displaystyle \frac{dz}{dt}$:
$\displaystyle \frac{dz}{dt}=\frac{-21z^3-36z}{6z^2+4}$
and for z=-2:
$\displaystyle \frac{dz}{dt}=\frac{-21(-1)^3-36(-1)}{6(-1)^2+4}=\frac{57}{10}=5.7$
I found it easier to set the function $\displaystyle F(x,y,z)=x^2z+xyz^3+6=0$ and use the chain rule:
$\displaystyle \frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}=(2xz+yz^3)(3t^2-3)+(xz^3)(6t-6)+(x^2+3xyz^2)\frac{dz}{dt}=0$
Solving for $\displaystyle \frac{dz}{dt}$ gives
$\displaystyle \frac{dz}{dt}=-\frac{(2xz+yz^3)(3t^2-3)+(xz^3)(6t-6)}{x^2+3xyz^2}$
At point (2,1,-1) and for t=2 the result is
$\displaystyle \frac{dz}{dt}=5.7$
which is the same as the above method.