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Exercise 17

Avg difficulty 2.5

Let z=f(x,y)=50−(x+1)2−(y+3)2z = f(x,y) = 50 - (x+1)^2 - (y+3)^2 and z=h(x,y)=24−2x−6yz = h(x,y) = 24 - 2x - 6y.

Suppose a person is walking on the surface z=f(x,y)z = f(x,y) in such a way that she walks the curve which is the intersection of ff and hh.

  1. Show that x(t)=4cos⁡(t)x(t) = 4 \cos(t) and y(t)=4sin⁡(t)y(t) = 4 \sin(t) is a parameterization of the “shadow” in the xyxy-plane of the curve that is the intersection of the graphs of ff and hh.

  2. Use the parameterization from part (a) to find the instantaneous rate at which her height is changing with respect to time at the instant t=2π/3t = 2\pi/3.

Long, Keller & Schlicker, Active Calculus Multivariable, §11.7 The Multivariable Chain Rule, Exercise 17

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Adapted from Active Calculus Multivariable © 2013–2026 Nicholas Long, Mitchel T. Keller, and Steven Schlicker. Licensed under CC BY-SA 4.0. Source: Active Calculus Multivariable. Imported at revision 692c03c.