h is continuous and h(3) < -5 < h(1), so there exists a value in that interval where h(x) = -5 by the Intermediate Value Theorem
b. h(3) - h(1)/3-1 = -10/2 = -5 and h is continuous and differentiable, so there exists a value c in that interval where h'(c) = -5 by the Mean Value Theorem.
b. dV/dt = 2000 - R(t), so dV/dt = 0 when R(t) = 2000
2000 = 400*sqrt(t), solving for t ---> t = 25 minutes. Since dV/dt > 0 when 0 < t < 25 and dV/dt < 0 when t > 25, the oil slick reaches its maximum volume 25 minutes after the device begins working
c. 60000 + integral from 0 to 25 of (2000-R(t)) dt
Comments
16^(1/3) * 3^(1/3) * x^2 * y = 48^(1/3) * x^2 * y
_______
√7-√5
12
_____ 3 is in the index of the sq root
√2
3√5x
_______
2x
(2√x+y) (-3√x-4√y)
............................................R
I used the S's for the integral symbol.
...∬〖√1-r^2 〗rdrdθ
.0..0
Is that converted correctly?
Find the surface area of r ⃑(u, v)=(u, v, 2uv) for the region R = {(u, v)| u^2 + v^2 ≤ 9}
h(3) = f(g(3)) - 6 = -7
h is continuous and h(3) < -5 < h(1), so there exists a value in that interval where h(x) = -5 by the Intermediate Value Theorem
b. h(3) - h(1)/3-1 = -10/2 = -5 and h is continuous and differentiable, so there exists a value c in that interval where h'(c) = -5 by the Mean Value Theorem.
c. w'(3) = f(g(3))*g'(3) = -1*2 = -2
d. g(1) = 2, so g^-1(2) = 1
(g^-1)(2) = 1/g'(g^-1(2)) = 1/g'(1) = 1/5
So your line is y -1 = (1/5)*(x-2)
dV/dt = 2pi*r*dr/dt + pi*(r^2)*dh/dt => 2000 = 2*pi*100*2.5 + pi*(100^2)*dh/dt. Solve for dh/dt
b. dV/dt = 2000 - R(t), so dV/dt = 0 when R(t) = 2000
2000 = 400*sqrt(t), solving for t ---> t = 25 minutes. Since dV/dt > 0 when 0 < t < 25 and dV/dt < 0 when t > 25, the oil slick reaches its maximum volume 25 minutes after the device begins working
c. 60000 + integral from 0 to 25 of (2000-R(t)) dt