Related Rates Section 4.6b.

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Presentation transcript:

Related Rates Section 4.6b

Do Now: #10, p.238 Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates: m/sec m/sec m/sec Find the rates at which the box’s (a) volume, (b) surface area, and (c) diagonal length are changing at the instant when x = 4, y = 3, and z = 2.

Do Now: #10, p.238 m/sec m/sec m/sec x = 4 m, y = 3 m, z = 2 m (a)

Do Now: #10, p.238 m/sec m/sec m/sec x = 4 m, y = 3 m, z = 2 m (b)

Do Now: #10, p.238 m/sec m/sec m/sec x = 4 m, y = 3 m, z = 2 m (c)

More Practice Problems A spherical balloon is inflated at the rate of How fast is the balloon’s radius increasing at the instant the radius is 5 ft? At the instant in question:

More Practice Problems A spherical balloon is inflated at the rate of (b) How fast is the surface area increasing at the same instant?

More Practice Problems A particle P(x, y) is moving in the coordinate plane in such a way that dx/dt = –1 m/sec and dy/dt = –5 m/sec. How fast is the particle’s distance from the origin changing as it passes through the point (5, 12)? P(5, 12) D Need y Based on the given information, should dD/dt be positive or negative? x

More Practice Problems A particle P(x, y) is moving in the coordinate plane in such a way that dx/dt = –1 m/sec and dy/dt = –5 m/sec. How fast is the particle’s distance from the origin changing as it passes through the point (5, 12)? P(5, 12) D y x At the moment in question, the particle is moving toward the origin at a rate of 5 m/sec

More Practice Problems Now let’s work through #30 on p.240 together… s = distance ball has fallen x = distance from bottom of pole to shadow Light Ball at t = 0 1/2 sec later 50-ft pole Need Shadow x 30 x(t)

More Practice Problems Now let’s work through #30 on p.240 together… Use similar triangles to relate s and x. Light Ball at t = 0 1/2 sec later 50-ft pole Shadow x 30 x(t)

More Practice Problems Now let’s work through #30 on p.240 together… Differentiate with respect to t. Light Ball at t = 0 1/2 sec later The shadow is moving at a velocity of –1500 ft/sec. 50-ft pole Shadow x 30 x(t)