Respuesta :
Answer:
[TeX] \frac{dT}{dt}= e^{sin t}[cos^{3}t- 2costsin t]+sin^{4}t-3cos^{2}tsin^{2}t[/TeX]
Step-By-Step Explanation:
Given [TeX] T(x, y) = x^{2}e^{y} – xy^{3} [/TeX] where x = cost, y = sin t.
[TeX]\frac{dx}{dt}=-sin t, \frac{dy}{dt}=cos t [/TeX]
Using Chain Rule
[TeX] \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}[/TeX]
Substituting the values
[TeX] x = cost, y = sin t, \frac{dx}{dt}=-sin t, \frac{dy}{dt}=cos t [/TeX]
[TeX] \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)[/TeX]
[TeX] \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)[/TeX]
[TeX] \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)[/TeX]
Therefore, the rate of change of the temperature, [TeX] \frac{dT}{dt} [/TeX] the moth feels is:
[TeX] \frac{dT}{dt}= e^{sin t}[cos^{3}t- 2costsin t]+sin^{4}t-3cos^{2}tsin^{2}t[/TeX]
