MATH SOLVE

7 months ago

Q:
# Find positive numbers a and b so that the change of variables s=ax,t=by transforms the integral ∫∫rdxdy into ∫∫t∣∣∣∂(x,y)∂(s,t)∣∣∣dsdt for the region r, the rectangle 0≤x≤15, 0≤y≤60 and the region t, the square 0≤s,t≤1. a= b= what is ∣∣∂(x,y)∂(s,t)∣∣ in this case? ∣∣∂(x,y)∂(s,t)∣∣=

Accepted Solution

A:

s = ax

It is required to find the value of a which make:

0 ≤ s ≤ 1 when 0 ≤ x ≤ 15

∴ s = 0 when x = 0 ⇒⇒⇒ a = 0 (unacceptable)

or s = 1 when x = 15 ⇒⇒⇒ a = s/x = 1/15

Similarly, t = by

It is required to find the value of b which make:

0 ≤ t ≤ 1 when 0 ≤ y ≤ 60

∴ t = 0 when y = 0 ⇒⇒⇒ b = 0 (unacceptable)

or t = 1 when y = 60 ⇒⇒⇒ b = t/y = 1/60

[tex]a = \frac{1}{15} \ \ \ and \ \ \ b= \frac{1}{60} [/tex]

∴ x = 15s and y = 60t

[tex]J ( \frac{x,y}{s,t}) = \left[\begin{array}{ccc} \frac{dx}{ds} & & \frac{dx}{dt} \\ & & \\ \frac{dy}{ds} & & \frac{dy}{dt} \end{array}\right] = \left[\begin{array}{ccc}15& &0\\ & & \\0& &60\end{array}\right] = 900[/tex]

Note: it is not matrix it is determinant

It is required to find the value of a which make:

0 ≤ s ≤ 1 when 0 ≤ x ≤ 15

∴ s = 0 when x = 0 ⇒⇒⇒ a = 0 (unacceptable)

or s = 1 when x = 15 ⇒⇒⇒ a = s/x = 1/15

Similarly, t = by

It is required to find the value of b which make:

0 ≤ t ≤ 1 when 0 ≤ y ≤ 60

∴ t = 0 when y = 0 ⇒⇒⇒ b = 0 (unacceptable)

or t = 1 when y = 60 ⇒⇒⇒ b = t/y = 1/60

[tex]a = \frac{1}{15} \ \ \ and \ \ \ b= \frac{1}{60} [/tex]

∴ x = 15s and y = 60t

[tex]J ( \frac{x,y}{s,t}) = \left[\begin{array}{ccc} \frac{dx}{ds} & & \frac{dx}{dt} \\ & & \\ \frac{dy}{ds} & & \frac{dy}{dt} \end{array}\right] = \left[\begin{array}{ccc}15& &0\\ & & \\0& &60\end{array}\right] = 900[/tex]

Note: it is not matrix it is determinant