Question

(4) Suppose a moth is flying in a circle about a candle flame so that its position at time t is given by x = cost, y = sin t. Suppose that the air temperature is given by T(x, y) = (x^2)(e^y) − xy3. Use the Chain Rule to find a formula for dT/dt, the rate of change of the temperature the moth feels.

Answer 1

Answer:

$\frac{dT}{dt}= e^{sin t}[cos^{3}t- 2costsin t]+sin^{4}t-3cos^{2}tsin^{2}t$

Step-By-Step Explanation:

Given $T(x, y) = x^{2}e^{y} – xy^{3}$ where x = cost, y = sin t.

$\frac{dx}{dt}=-sin t, \frac{dy}{dt}=cos t$

Using Chain Rule

$\frac{dT}{dt}= x^{2}e^{y}\frac{dy}{dt}+ 2xe^{y}\frac{dy}{dt}-y^{3}\frac{dx}{dt}-x3y^{2}\frac{dy}{dt}$

Substituting the values  

$x = cost, y = sin t, \frac{dx}{dt}=-sin t, \frac{dy}{dt}=cos t$

$\frac{dT}{dt}= (cost)^{2}e^{sin t}(cos t)+ 2(cost)e^{sin t}(-sin t)-(sin t)^{3}(-sin t)-(cost )3(sin t)^{2}(cos t)$

$\frac{dT}{dt}= cos^{2}te^{sin t}(cos t)+ 2(cost)e^{sin t}(-sin t)-(sin^{3}t)(-sin t)-(cost )3(sin^{2}t) (cos t)$

$\frac{dT}{dt}= cos^{3}te^{sin t}- 2(cost) (sin t)e^{sin t}+(sin^{4}t)-(cos^{2}t)3(sin^{2}t)$

Therefore, the rate of change of the temperature, $\frac{dT}{dt}$ the moth feels is:

$\frac{dT}{dt}= e^{sin t}[cos^{3}t- 2costsin t]+sin^{4}t-3cos^{2}tsin^{2}t$