Rate of change in temperature as a bug moves. My question comes from a problem from a multivariable calculus class. The problem states 
1
.
The temperature on a hot surface is given by
$T=100e^{−(x^2+y^2)}$. A bug follows the trajectory
$r(t) =<t\cos(2t), t\sin(2t)>$
.
What
is
the
rate
that
temperature
is
changing
as
the
bug
moves?
using the chain rule in respects to $\frac {dT}{dt}$ then substituting t for x and y I get
$-200t\cos(2t)e^{-((t\cos(2t))^2+(t\sin(2t)^2)}(\cos(2t)-2t\sin(2t))-200t\sin(2t)e^{-((t\cos(2t))^2+(t\sin(2t)^2)}(\sin(2t)+2t\cos(2t)$ 
but it is not clear how to reduce to the answer:
 $-200te^{-t^2}$
 A: We have: $T(x(t),y(t)) = 100e^{-t^2} \to T'(t) = -200te^{-t^2}$
A: Hint
The key is to use $\sin^2(2t)+\cos^2(2t)=1$. 
So, simplify first the exponential term, factor it and expand the remaining. You will arrive to the correct result.
A: So the temperature at a point $(x, y)$ is $T(x, y) = 100 e^{-(x^2 + y^2)}$. Your path $r$ is parametrised by $t$ in that
$x(t) = t \cos(2t)$
and
$y(t) = t \sin(2t)$
so we can find the temerpature along the path $r$ just by 
$T(x(t), y(t)) = 100 e^{-(t^2 \cos^2(2t) + t^2 \sin^2(2t))} = 100 e^{-t^2}$,
since $\cos^2 + \sin^2 = 1$.
Note now the temperature along the path only depends on $t$, so we can differentiate:
$\frac{\mathrm{d}T}{\mathrm{d}t} = 100e^{-t^2} \times (-2t) = -200 e^{-t^2}$,
by an appliaction of the chain rule.
A: You made a mistake in your computation. An approach could be to write
$$
\frac {dT}{dt}=\frac {\partial x}{\partial t}\frac {\partial T}{\partial x}+\frac {\partial y}{\partial t}\frac {\partial T}{\partial y} \tag1
$$ with
$$
\frac {\partial x}{\partial t}=\cos(2t)-2t\sin(2t), \quad \frac {\partial y}{\partial t}=\sin(2t)+2t\cos(2t) \tag2
$$ and
$$
\frac {\partial T}{\partial x}=-2x\:T, \quad \frac {\partial T}{\partial y}=-2y\:T \tag3
$$ then plugging $(2)$ and $(3)$ in $(1)$, you get
$$
\begin{align}
\frac {dT}{dt}&=(\cos(2t)-2t\sin(2t))(-2x\:T)+(\sin(2t)+2t\cos(2t))(-2y\:T)\\\\
&=(\cos(2t)-2t\sin(2t))(-2t\cos(2t)\:T)+(\sin(2t)+2t\cos(2t))(-2t\sin(2t)\:T)\\\\
&=[(\cos(2t)-2t\sin(2t))\cos(2t)+(\sin(2t)+2t\cos(2t))\sin(2t)](-2t\:T)\\\\
&=[(\cos^2(2t)+\sin^2(2t))-2t\sin(2t)\cos(2t)+2t\cos(2t)\sin(2t)](-2t\:T)\\\\
&=-2t\:T\\\\
&=-200te^{-t^2}
\end{align}
$$
giving the desired result.
