Question

4. An airplane flying horizontally at an altitude of y = 3 km and at a speed of
480 km/h passes directly above an observer on the ground. How fast is the
distance D from the observer to the airplane increasing 30 seconds later?

5. A kite is rising vertically at a constant speed of 2 m/s from a location at
ground level which is 8 m away from the person handling the string of the
kite
(a) Let z be the distance from the kite to the person. Find the rate of change
of z with respect to time t when z = 10.
(b) Let x be the angle the string makes with the horizontal. Find the rate
of change of x with respect to time t when the kite is y = 6 m above
ground.

6. A balloon is rising at a constant speed 4m/sec. A boy is cycling along a
straight road at a speed of 8m/sec. When he passes under the balloon, it is
36 metres above him. How fast is the distance between the boy and balloon
increasing 3 seconds later.

Answer 1

Answer on Question #64599 – Math – Calculus

Question

4. An airplane flying horizontally at an altitude of $y = 3$ km and at a speed of $480$ km/h passes directly above an observer on the ground. How fast is the distance $D$ from the observer to the airplane increasing 30 seconds later?

Solution

$D$ is the distance from the airplane to the observer and $x$ is the (horizontal) distance traveled by the airplane from the moment it passed over the observer.

We know that $v = dx/dt = 480 \, \text{km/h}$.

We want to know $dD/dt$ 30 seconds after the plane flew over the observer.

$$t = 30 \sec = 30 \cdot \frac{1}{3600} \text{ hours} = \frac{1}{120} \text{ hours}$$

Applying the Pythagorean Theorem to a right triangle

$$D^2 = x^2 + y^2, \quad D^2 = x^2 + 3^2$$

Taking derivatives with respect to time $t$ on both sides we get

$$2D \frac{dD}{dt} = 2x \frac{dx}{dt},$$

so that

$$\frac{dD}{dt} = \frac{x}{D} \cdot \frac{dx}{dt}$$

We have that

$$t = 30 \sec = \frac{1}{120} \text{ hours}, \quad \frac{dx}{dt} = v = 480 \frac{\text{km}}{\text{h}}, \quad x = vt = 480 \cdot \frac{1}{120} = 4 (\text{km}),$$ $$D^2 = 4^2 + 3^2 = 25 (\text{km}^2), \quad D = 5 \text{ km}.$$

Then

$$\frac{dD}{dt} = \frac{4}{5} \cdot 480 = 384 \left(\frac{\text{km}}{\text{h}}\right).$$

Answer: $384 \frac{\text{km}}{\text{h}}$.

Question

5. A kite is rising vertically at a constant speed of $2$ m/s from a location at ground level which is $8$ m away from the person handling the string of the kite.

(a) Let $z$ be the distance from the kite to the person. Find the rate of change

of $z$ with respect to time $t$ when $z = 10$ .

(b) Let $x$ be the angle the string makes with the horizontal. Find the rate of change of $x$ with respect to time $t$ when the kite is $y = 6 \, \text{m}$ above ground.

Solution

(a) Applying the Pythagorean theorem to a right triangle

$$z ^ {2} = 8 ^ {2} + y ^ {2}$$

We know that

$$v _ {y} = \frac {d y}{d t} = 2 \frac {m}{s}, y = v _ {y} t.$$

Then

$$z ^ {2} = 6 4 + \left(v _ {y} t\right) ^ {2}$$

Taking derivatives with respect to time $t$ on both sides we get

$$2 z \frac {d z}{d t} = 2 v _ {y} ^ {2} t.$$

So that

$$\frac {d z}{d t} = \frac {1}{z} v _ {y} ^ {2} t = \frac {y}{z} v _ {y}.$$

We have that

$$z = 1 0 m, 1 0 ^ {2} = 6 4 + y ^ {2}, y = 6 m.$$

Therefore

$$\frac {d z}{d t} = \frac {6}{1 0} \cdot 2 = 1. 2 \left(\frac {m}{s}\right).$$

(b) Using the definition

$$\tan x = \frac {y}{8}$$

Taking derivatives with respect to time $t$ on both sides we get

$$\frac {1}{\cos^ {2} x} \cdot \frac {d x}{d t} = \frac {1}{8} \cdot \frac {d y}{d t}.$$

So that

$$\frac {d x}{d t} = \frac {1}{8} \cdot \cos^ {2} x \cdot v _ {y}.$$

We have that

$$y = 6 \, m, \tan x = \frac{6}{8} = \frac{3}{4}, 1 + \tan^2 x = 1 + \left(\frac{3}{4}\right)^2 = \frac{25}{16} = \frac{1}{\cos^2 x}, \cos^2 x = \frac{16}{25}.$$

Therefore

$$\frac{dx}{dt} = \frac{1}{8} \cdot \frac{16}{25} \cdot 2 = \frac{4}{25} = 0.16 \, \frac{rad}{s}.$$

Answer: (a) $1.2 \, \frac{m}{s}$; (b) $0.16 \, \frac{rad}{s}$.

Question

6. A balloon is rising at a constant speed $4\,\mathrm{m/sec}$. A boy is cycling along a straight road at a speed of $8\,\mathrm{m/sec}$. When he passes under the balloon, it is 36 metres above him. How fast is the distance between the boy and balloon increasing 3 seconds later?

Solution

D is the distance between the boy and balloon, $x$ is the (horizontal) distance traveled by the boy from the moment it passed under the balloon and $y$ is the altitude of the balloon.

Applying the Pythagorean theorem to a right triangle

$$D^2 = x^2 + y^2$$

We know that

$$v_x = \frac{dx}{dt} = 8 \, \frac{m}{sec}, \quad v_y = \frac{dy}{dt} = 4 \, \frac{m}{sec}, \quad x = v_xt, \quad y = y_0 + v_yt, \quad y_0 = 36 \, m.$$

Then

$$D^2 = (v_xt)^2 + (y_0 + v_yt)^2$$

Taking derivatives (with respect to time, $t$) on both sides we get

$$2D \frac{dD}{dt} = 2 v_x^2 t + 2 v_y (y_0 + v_yt),$$

so that

$$\frac {d D}{d t} = \frac {1}{D} \cdot \left(\mathrm {v} _ {x} ^ {2} t + \mathrm {v} _ {y} \left(y _ {0} + \mathrm {v} _ {y} t\right)\right)$$

We have that

$$t = 3 \sec , D ^ {2} = (8 \cdot 3) ^ {2} + (3 6 + 4 \cdot 3) ^ {2} = 2 8 8 0 (m ^ {2}), D = 2 4 \sqrt {5} m.$$

Therefore

$$\frac {d D}{d t} = \frac {1}{2 4 \sqrt {5}} \left(8 ^ {2} \cdot 3 + 4 (3 6 + 4 \cdot 3)\right) = \frac {1 6}{\sqrt {5}} = \frac {1 6 \sqrt {5}}{5} \left(\frac {m}{s e c}\right).$$

Answer: $\frac{16\sqrt{5}}{5}\frac{m}{sec}$ .

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