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

To find the rate of change of the temperature the moth feels, we need to use the Chain Rule. So, the rate of change of the temperature the moth feels, is: -2cos(t)eˢᶦⁿ ᵗ)sin(t) + sin(t)⁴ + cos(t)²eˢᶦⁿ ᵗ)cos(t) - 3cos(t)sin(t)³.

Step-by-step explanation:

To find the rate of change of the temperature the moth feels, we can use the Chain Rule. The Chain Rule states that if z = f(x, y) and x = g(t) and y = h(t), then dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt).

In this case, we have T(x, y) = (x²)(eʸ) − xy³, where x = cos(t) and y = sin(t).

We want to find dT/dt.

First, let’s find ∂T/∂x and ∂T/∂y:

∂T/∂x = 2xe² - y³

∂T/∂y = x²eʸ - 3xy²

Now, let’s find dx/dt and dy/dt:

dx/dt = -sin(t) dy/dt = cos(t)

Using the Chain Rule, we can find dT/dt:

dT/dt = (∂T/∂x)(dx/dt) + (∂T/∂y)(dy/dt) = (2xeʸ - y³)(-sin(t)) + (x²eʸ - 3xy³)(cos(t))

Substituting x = cos(t) and y = sin(t):

= (2cos(t)eˢᶦⁿ ᵗ) - sin(t)³)(-sin(t)) + (cos(t)²eˢᶦⁿ ᵗ) - 3cos(t)sin(t)²)(cos(t))

Simplifying this expression gives us the final formula for dT/dt.

Final Answer:

The formula for dT/dt, the rate of change of the temperature the moth feels, is:

-2cos(t)eˢᶦⁿ ᵗ)sin(t) + sin(t)⁴ + cos(t)²eˢᶦⁿ ᵗ)cos(t) - 3cos(t)sin(t)³

This represents the instantaneous rate of change of the temperature experienced by the moth as it flies in a circular path around the candle flame.

Answer 2

[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]