Almost done! Quick points! I just need help finishing the last step on these two

Question

Almost done! Quick points! I just need help finishing the last step on these two problems!!! Points will be given upon solution!

A hot-air balloon is 120 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 45 mi/hr (66 ft/ s). If the balloon rises vertically at a rate of 10 ft / s, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later? Let x be the horizontal distance from the balloon to the motorcycle, y be the vertical distance from the balloon to the road, and z be the distance between the motorcycle and the balloon. Write an equation relating x, y, and z. X^2+ y^2 = z^2 Find the related rates equation. The rate of change of the distance between the motorcycle and the balloon after 10 seconds is about ft/s. (Round to two decimal places as needed.) The hands of a clock in some tower are approximately 4.5 m and 4 m in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the law of cosines.) Write an equation relating the angle between the two clock hands, 6, and the distance between the tips of the two hands, c. C^2=36.25-36cos theta Find the related rates equation. The distance between the tips of the hands is changing at a rate of m/hr at 9:00. (Round to two decimal places as needed.)

Explanation / Answer

hello user hope this helps your querry

If we let y be the angle between the two hands and x be the distance between the two tips, then, by the law of cosines, we have:
x^2 = 4^2 + 4.5^2 - 2*4*4.5cos(y)
x^2 = 36.25 - 36cos(y)

Take the derivative of both sides with respect to t, time.
2x*dx/dt = 36sin(y)*dy/dt

Since it is 9:00, the angle between the two hands must be y = ?/2. And since there is a right triangle, x = ?(4^2 + 4.5^2) = ?36.25. In order to find dy/dt, consider the fact that the hours hand goes around 2? in one hour and the minutes hand goes around 2? in 1/60 hour, therefore we have dy/dt = 2? - 2?/(1/60) = -118?. Plug that all in:
2(?36.25)*dx/dt = 36sin(?/2)*(-118?)
2(?36.25)*dx/dt = 36(1)*(-118?)
2(?36.25)*dx/dt = -4248?
(?36.25)*dx/dt = -2124?

dx/dt = -2124/?36.25 ? -352.78

So the distance ischanging at the rate of 352.78 meters per hour.

2)

Let

x - be the distance the motarcycle moves, from the point it was directly under the balloon
y - be the distance the balloon rose up, starting from the time the motarcycle went through it
z - be the distance between the balloon and the motarcycle

If you try to draw the representation of the situation above, you'll come up with a right triangle representation. Thus, with the above variables to represent the important distances, we can relate all three with the use of Pythagorean Theorem

x^2 + (y + 120)^2 = z^2

Since y is measured from the time that the motarcycle went under it, we have to add the 120 ft, in order to have the correct dimensions of the right triangle. Taking the derivative with respect to time, we have

2x (dx/dt) + 2(y + 120) (dy/dt) = 2z (dz/dt)

Note that all distances are changing, thus we have rates for all 3 distances. In the above,

dx/dt - refers to the rate x is changing, or how fast is the motarcycle  moving = 66 ft/s
dy/dt - refers to the rate y is changing, or how fast is the balloon rising = 10 ft/s
dz/dt - refers to the rate z is changing, or how fast are the motarcycle  and balloon separating = ?

By the way, dx/dt, and dy/dt are both positive, because they refer to a rate which INCREASES the values of x, and y, respectively. Sign conventions are very important in related rates problems.

We know what dx/dt, and dy/dt are, and we are looking for dz/dt (as the problem is asking us to). The problem is, in our equation above, we need to determine what x, y, and z are, of what those distances are, 10s after the motarcycle  passed under the balloon.

Since we know the speeds, then after 10 s,

x = 10 * 66 = 660 ft
y = 10 * 10 = 100 ft

And for z,

x^2 + (y + 120)^2 = z^2
(660)^2 + (100+ 120)^2 = z^2
435600 + 48400 = z^2
z = sqrt(92000)
z = 303.3

Thus, plugging them all in to our MAIN equation, we have

2x (dx/dt) + 2(y + 120) (dy/dt) = 2z (dz/dt)
2(660) (66) + 2(120 + 100) (10) = 2(303.3) (dz/dt)
87120+4400 = 303.3dz/dt
dz/dt = 301.4 ft/s

hence the rat of change of motarcycle after 10 sec is 301.74 ft /sec

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