Final answer:
The second partial derivatives of the function T = e−8 cos(θ) are calculated with respect to r and θ, resulting in T_{rr}, T_{θθ}, and the mixed partials T_{rθ} = T_{θr}, providing insights into the function's curvature.
Step-by-step explanation:
To find all the second partial derivatives of the function T = e−8 cos(θ), we need to differentiate twice with respect to both r and θ. Since T is a function of both r and θ, we will have partial derivatives with respect to r (written as T_{rr}), θ (written as T_{θθ}), and the mixed partial derivatives T_{rθ} and T_{θr}.The pure second partial derivative with respect to r is:T_{rr} = e−8(64 cos(θ))The pure second partial derivative with respect to θ is:T_{θθ} = -e−8 cos(θ)The mixed partial derivative first with respect to r and then θ is:T_{rθ} = e−8(8 sin(θ))The mixed partial derivative first with respect to θ and then r is:T_{θr} = e−8(8 sin(θ)
)Note that T_{rθ} and T_{θr} are equal, which is a result of Clairaut's theorem on the equality of mixed partials. To write the final answer in 20 words: The second partial derivatives are T_{rr} = 64e−8 cos(θ), T_{θθ} = -e−8 cos(θ), and T_{rθ} = T_{θr} = 8e−8 sin(θ).To find all the second partial derivatives of T = e-8rcos(θ), we need to take the partial derivatives with respect to both r and θ twice. First, we differentiate with respect to rdT/dr = e-8rcos(θ) * (-8) = -8e-8rcos(θ)Then, we differentiate again with respect to r:d²T/dr² = (-8) * (-8e-8rcos(θ)) = 64e-8rcos(θ)Next, we differentiate with respect to θ:dT/dθ = -e-8rsin(θ)Finally, we differentiate again with respect to θ: