Related Rates, Linear Approximation and Differentials - Solved Problems | M 408C, Study notes of Mathematics

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M408C: Related Rates, Linear Approximation, &

Differentials

September 30, 2008

4. (3.8.17) A man starts walking north at 4 ft/s from a point P. Five minutes later, a woman starts

walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving

apart 15 minutes after the woman starts walking?

Solution: Let x = x(t), y = y(t) denote the distance travelled by the man and woman, re-

spectively. Thus dx/dt = 4 ft/s and dy/dt = 5 ft/s. Let z = z(t) denote the distance between

them, which is the hypotenuse of the right triangle with height x + y and base 500 ft. Thus

z^2 = (x + y)^2 + 500^2. Differentiating this with respect to the independent variable t (via im-

plicit differentiation), we get 2z

dz dt = 2(x^ +^ y)

dx dt +^

dy dt

. We want dz/dt since that is the rate

at which the man and woman are walking apart. We know dx/dt and dy/dt, so we need to

determine x, y, and z. Now, 15 minutes after the woman starts walking, the man has been walk-

ing for 20 min. So x = 4

f t s ·^20

min 1 ·^60

s min = 4800 ft.^ After 15 minutes, the woman has walked

y = 5

f t s ·^20

min 1 ·^60

s min = 4500 ft. Plugging this in for^ z^ =^

(x + y)^2 + 500^2 , we get z =

ft after 15 minutes. Solving for dz/dt, we get

dz

dt

x + y

z

dx

dt

dy

dt

≈ 8. 99 f t/s.

5. (3.9.28) Use linear approximation to estimate

Solution: Set f (x) =

x so that we want to a linear approximation to f at a = 100. We quickly

compute f ′(x) = 1 2

√ x

, so

L(x) = f (a) + f ′(a)(x − a) =

(x − 100) = 10 +

x − 100

20

Plugging in x = 99.8 = 99

4 5 , we get^

99. 8 ≈ L(99.8) = 10 +

− (^15) 20 = 10^ −^

1 100 =^

999 100 = 9.^99.

6. (3.9.36) One side of a right triangle is known to be 20 cm long and the opposite angle is measured

as 30◦, with a possible error of ± 1 ◦. Use differentials to estimate the error in computing the length

of the hypotenuse. What is the percent error?

Solution: Let θ be the angle in question, and let y(θ) be the length of the hypotenuse. Thus

sin θ = (^) y^20 (θ) , so y(θ) = 20/ sin θ and set y = f (θ) wher f (θ) = 20/ sin θ. We are told the error is

± 1 ◦, so dθ = ± 1 ◦^ = ± 180 π radians. Now, to determine the error in computing the length, we need

to compute dy = f ′(θ)dθ:

dy =

[

20 · (−(sin θ)−^2 ) · cos θ

]

[

π

180

]

π cos θ

9(sin θ)^2

Evaluating this at θ = 30◦^ = π/6, we get

dy = ±

π

√ 3 2 9(1/2)^2

π ≈ ± 1 .21 cm

. The percent error is given by (recall y = 20/ sin θ)

dy

y

2

√ 3 9 π 20 1 / 2

π ·

π ≈ ± 0 .03 = ±3%.