SUNY Geneseo Department of Mathematics

Related Rates, Part 2

Thursday, March 14

Math 221 03
Spring 2019
Prof. Doug Baldwin

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Questions?

More Related Rates Examples

Pedestrian

The problem we started yesterday, with a pedestrian and a car approaching a crosswalk. We wanted to know what dz/dt is, i.e., how fast the straight-line distance between the pedestrian and the car is changing. That distance is given by the Pythagorean Theorem.

Pedestrian's and car's paths to crosswalk forming a right triangle

We came up with two ways to find dz/dt. the first solves for z in terms of x and y and then differentiates, which is straightforward but requires using the power rule with a fractional exponent, which technically we have never proven works. Nonetheless, completing that solution by plugging in numbers for x, y, and their derivatives gives an answer:

Differentiate square root of x squared plus y squared and plug in numbers to get about 22.16 feet per second

The other approach is to implicitly differentiate the equation for z2 in terms of x2 and y2, which is subtler but doesn’t rely on any unproven math. That method produces the same answer:

Differentiate x squared plus y squared equals z squared and plug in numbers to get about 22.16 feet per second

Hiding Spy

Super spy Jane Bond is hiding under the wall to Dr. Doom’s estate, trying to avoid the estate’s security camera. The camera is on a telescoping pole. How fast does the region below the wall where Jane Bond is safe from the camera change as the camera’s height changes?

Spy beside wall with camera on pole on other side

The key idea for connecting the safe region to the camera’s height is to realize that there are 2 similar triangles in this problem: one defined by the safe region and the wall, and the other defined by the safe region plus the pole’s distance inside the wall and the camera’s height:

Triangle involving pole, ground and camera sight line vs triangle involving wall, ground, and sight line

Because the triangles are similar, ratios of corresponding sides are equal:

x over w equals the quantity x + y over h

Now we can rewrite this equation to give x in terms of h (and w and y, which are constants):

x over w equals x plus y over h rewrites to x equals w times y over the quantity h minus w

And finally we can take derivatives:

Derivative of x equals negative w times y over quantity h minus w squared all times derivative of h

Problem Set

See handout for details.

Next

Approximation with derivatives.

Read section 4.2

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