The fallout, in kilograms per square kilometer, from a volcano eruption is given by V(s , t)= √ t e^ -s where s = distance from the volcano and t = time.
∂^2. V / ∂. t^2=
∂.^2 V / ∂. s ∂. t =
thank you
∂^2. V / ∂. t^2=
∂.^2 V / ∂. s ∂. t =
thank you
-
∂V/∂t = (1/2)t^(-1/2) e^(-s)
Differentiate again with respect to t to find
∂²V/∂t² = (-1/4)t^(-3/2) e^(-s)
Now differentiate ∂V/∂t with respect to s to find
∂²V/∂s∂t = -(1/2)t^(-1/2) e^(-s)
Differentiate again with respect to t to find
∂²V/∂t² = (-1/4)t^(-3/2) e^(-s)
Now differentiate ∂V/∂t with respect to s to find
∂²V/∂s∂t = -(1/2)t^(-1/2) e^(-s)