The radius of a right circular cone is increasing at a rate of 4 inches per seco

Question

The radius of a right circular cone is increasing at a rate of 4 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 20 inches and the height is 20 inches? Answer in cubic inches per second.

Explanation / Answer

1) ?z = ==> ?z(2, 1) = . So, the equation of the tangent plane is z - (-27) = -36(x - 2) + -18(y - 1). 2) This is done just like 1). 3) Just find the equation of the tangent plane at (x, y, z) = (-8, 4, -16) [as above] and solve for z. 4) Here, A(x,y,z) = 2xy + 2xz + 2yz. Letting x = 80, y = 50, z = 100, we see that ?x = ?y = ?z = 0.2 dA = ?A · = (2y + 2z) dx + (2x + 2z) dy + (2x + 2y) dz. Letting dx = ?x, dy = ?y, dz = ?z, we have that ?A ˜ dA. Hence, ?A ˜ (2y + 2z) ?x + (2x + 2z) ?y + (2x + 2y) ?z = (2 * 50 + 2 * 100) (0.2) + (2 * 80 + 2 * 100) (0.2) + (2 * 80 + 2 * 50) (0.2). = 184 sq. cm. 5) This is a related rates problem! Use V = (p/3) r^2 h. Moreover, dr/dt = 3, dh/dt = -5 Differentiating with respect to t, dV/dt = (2p/3) r dr/dt * h + (p/3) r^2 dh/dt. When r = 50, h = 20, we have dV/dt = (2p/3) * 50 * 3 * 20 + (p/3) * 50^2 * (-5) = -6500p/3. That is, the volume is decreasing at a rate of 6500p/3 cubic inches.

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