Q. 35 - Calculus

Question

Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers. 

$\frac{d\rho }{dt} \mathrm{when} \rho =\sqrt{x^2+y^2+z^2}, x=\sqrt{t}, y=t^2, z=t^3$

Step-by-Step Solution

Answer

The value of $\frac{d\rho }{dt}=\frac{1}{2}\left(1+4t^3+3t^5\right){\left(t+t^4+t^6\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Step 1. Given Information.

$$\begin{gathered} \rho =\sqrt{x^2+y^2+z^2} \\ x=\sqrt{t} \\ y=t^2 \\ z=t^3 \end{gathered}$$

Step 2. Calculation.

By Theorem 12.34, we have 

$\frac{d\rho }{dt}=\frac{\partial \rho }{\partial x}·\frac{dx}{dt}+\frac{\partial \rho }{\partial y}·\frac{dy}{dt}+\frac{\partial \rho }{\partial z}·\frac{dz}{dt}------\left(1\right)$

So first we find 

$\frac{\partial \rho }{\partial x},\frac{dx}{dt},\frac{\partial \rho }{\partial y},\frac{dy}{dt},\frac{\partial \rho }{\partial z},\frac{dz}{dt}$

So we have 

$$\begin{gathered} \frac{\partial \rho }{\partial x}=\frac{1·2x}{2\sqrt{x^2+y^2+z^2}}=\frac{x}{\sqrt{x^2+y^2+z^2}} \\ \frac{\partial \rho }{\partial y}=\frac{1·2y}{2\sqrt{x^2+y^2+z^2}}=\frac{y}{\sqrt{x^2+y^2+z^2}} \\ \frac{\partial \rho }{\partial z}=\frac{1·2z}{2\sqrt{x^2+y^2+z^2}}=\frac{z}{\sqrt{x^2+y^2+z^2}} \\ \frac{dx}{dt}=\frac{1}{2t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \frac{dy}{dt}=2t \\ \frac{dz}{dt}=3t^2 \end{gathered}$$

Step 3. Calculation.

Use these above values in (1) we get, 

$$\begin{gathered} \frac{d\rho }{dt}=\frac{\partial \rho }{\partial x}·\frac{dx}{dt}+\frac{\partial \rho }{\partial y}·\frac{dy}{dt}+\frac{\partial \rho }{\partial z}·\frac{dz}{dt} \\ \frac{d\rho }{dt}=\frac{x}{2t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{2yt}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{3t^{2}z}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

So from here, putting the value of $x,y,z$in terms of $t$ we get

$$\begin{gathered} \frac{d\rho }{dt}=\frac{x}{2t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{2yt}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{3t^{2}z}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \frac{d\rho }{dt}=\frac{1}{\sqrt{t+t^4+t^6}}\left(\frac{1}{2}+2t^3+3t^5\right) \\ \frac{d\rho }{dt}=\frac{1}{2}\left(1+4t^3+3t^5\right){\left(t+t^4+t^6\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Step 4. Conclusion.

The value of $\frac{d\rho }{dt}=\frac{1}{2}\left(1+4t^3+3t^5\right){\left(t+t^4+t^6\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.