Group+F+Period+8


 * Section 4.6:****# 6, 12, and 18**

-- Marshall Hart, Aaron Walters, and Nicolas Ciccone -- 6. A = (1/2) absinΘ dA/dt = (1/2) (abcosΘ(dΘ/dt)+asinΘ(db/dt)+bsinΘ(da/dt))

The problem asks, if a and b are the lengths of the sides of a triangle and Θ is the included angle in the given area above, to show how dA/dt relates to da/dt, db/dt, and dΘ/dt. The solution is simply an extension of the product rule with three differentials.

12. V = (4/3)π(r^3) (dV/dt) = 4π(r^2)(dr/dt) A = 4π(r^2) dA/dt = 8πr(dr/dt) dr/dt = dA / (dt8πr) dr/dt = dV / (dt4π(r^2)) dA / (dt8πr)) = dV / (dt4π(r^2)) dA / (8πr) = dV / (4π(r^2)) dV/dA = 4π(r^2) / 8πr dV/dA = (1/2)r k=1/2

A perfectly spherical droplet of mist is said to increase its moisture, or volume, at a rate proportional to its surface area. This implies that the radius, r, must increase at a constant rate, k. To start the problem, the volume, V, of any sphere must be (4/3)π(r^3), and the area, A, 4π(r^2). If we take the derivatives of both of these functions, we can find dV/dt and dA/dt each with respect to dr/dt. After a little algebra, we can find the values of dr/dt with respect to dA or dV, and set the two equal to each other. dt cancels out, and by order of operations, we can set dV/dA (proportional, as per the instructions) equal to 4π(r^2) / 8πr. Factoring leaves (1/2)r, so not only does this prove that r increases at a constant rate, but also that the constant rate, k, is 1/2.

18.) Water flows at a rate of

a. V = ( p /3)(y^2)(3R-y) V = ( p /3)(y^2)(39-y) y=8 m dV/dt=-6 m3/min dV/dt = ( p /3)((y^2)(-dy/dt) + (39-y)(2y(dy/dt)) -6 = ( p /3) ((8^2)(-dy/dt) + (39-8)(2y(dy/dt)) -6 = ( p /3) (-64(dy/dt) + 496(dy/dt)) -18/ p = -64(dy/dt) + 496(dy/dt) -18/ p = 432(dy/dt) dy/dt = (-18/ p ) / 432 dy/dt ≈ -0.01326 m/min

b. x=13-y x^2+r^2=169 (13-y)^2+r^2=169 169-26y+y^2+r^2=169 -26y+y^2+r^2=0 r^2 = 26y-y^2 r= ( 26y-y2)^1/2

c. dr/dt = (1/2) ((26y-y2)^-1/2) (26-2y) (dy/dt) dr/dt = (1/2) ((26(8)-82)^-1/2) (26-2(8)) ((-18/ p ) / 432) dr/dt = -0.0055262 m/min

Water flows at a rate of 6 m3/min from a hemispherical bowl-shaped reservoir of 13 m radius.

If the volume of the sphere of radius R is V = ( p /3)y2(3R-y) when the water is y units deep, we will first try to find the rate the water level is changing when the water is 8 m deep.

First, when taking the derivative of the given volume function, we are left with ( p /3)((y^2)(-dy/dt) + (39-y)(2y(dy/dt)); 3R, or 39, does not have a derivative since it is a constant. Since we are left with dR/dt with respect to dy/dt, the deepness of the water, we substitute 8 for y, and -6 for dV/dt, since the water is flowing out. After a little algebra, we are left with -18/ p = 432(dy/dt). We divide -18/ p by 432, and are left with -0.01326 m/min.

Second, to find the radius r of the water’s surface when the water is 3 m deep, we use Pythagorean Theorem to show that x^2+r^2=169, because 13^2 is 169. Let us substitute an arbitrary value of x for the space above the water level. Since the radius is constant throughout the hemisphere, 13=x+y, so x=13-y. We can substitute 13-y for x in our original model, and get (13-y)^2+r^2=169. Therefore, the value of r at y meters is (26y-y^2)^1/2.

Third, to show that the radius r is changing at 8 m deep, we may use the formula above, r= (26y-y2)^1/2, and find the derivative to show dr/dt with respect to dy/dt. To solve for dr/dt, we can substitute y=8 and dy/dt = -18/ p / 432. After the algebra, we are left with dr/dt = -0.0055262 m/min as our answer.