Question

the temperature at a point $(x,y)$ is $t(x,y)$, measured in degrees celsius. a bug crawls so that its position after $t$ seconds is given by $x = sqrt{3 + t}, y = 1 + \frac{1}{6}t$, where $x$ and $y$ are measured in centimeters. the temperature function satisfies $t_x(3,2)=6$ and $t_y(3,2)=5$. how fast is the temperature rising on the bugs path after 6 seconds? (round your answer to two decimal places.)

Explanation:

Step1: Find \(x\) and \(y\) at \(t = 6\)

When \(t=6\), \(x=\sqrt{3 + 6}=\sqrt{9}=3\), \(y=1+\frac{1}{6}\times6=2\)

Step2: Differentiate \(x\) and \(y\) with respect to \(t\)

Differentiate \(x=\sqrt{3 + t}=(3 + t)^{\frac{1}{2}}\) using the power - rule \((u^n)^\prime=nu^{n - 1}u^\prime\). Let \(u = 3 + t\), then \(x^\prime=\frac{1}{2}(3 + t)^{-\frac{1}{2}}\times1=\frac{1}{2\sqrt{3 + t}}\)
Differentiate \(y = 1+\frac{1}{6}t\), then \(y^\prime=\frac{1}{6}\)

Step3: Use the chain - rule for partial derivatives

The chain - rule for \(T(x(t),y(t))\) is \(\frac{dT}{dt}=T_x(x,y)\frac{dx}{dt}+T_y(x,y)\frac{dy}{dt}\)
When \(t = 6\), \(x = 3\), \(y = 2\), \(\frac{dx}{dt}\big|_{t = 6}=\frac{1}{2\sqrt{3+6}}=\frac{1}{6}\), \(\frac{dy}{dt}=\frac{1}{6}\)
Substitute \(T_x(3,2) = 6\), \(T_y(3,2)=5\), \(\frac{dx}{dt}=\frac{1}{6}\), \(\frac{dy}{dt}=\frac{1}{6}\) into the chain - rule formula:
\(\frac{dT}{dt}=T_x(3,2)\frac{dx}{dt}+T_y(3,2)\frac{dy}{dt}\)
\(\frac{dT}{dt}=6\times\frac{1}{6}+5\times\frac{1}{6}\)
\(\frac{dT}{dt}=\frac{6 + 5}{6}=\frac{11}{6}\approx1.83\)

Answer:

\(1.83\)