Plz help me with this questions!

[merve]

1) The minute hand of a clock is 8 centimeters long, whereas the hour hand is 4 centimeters long. How fast is the distance between the tips of the hands changing at the 11 o'clock position? (Your answer should be in units of cm/hr.)

2)A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?

3) A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 9 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 16 ft/min, at what rate will the boat be approaching the dock when 110 ft of rope is out?

The boat will be approaching the dock at _____________ ft/min.

Note: There is a diagram of this situation with Problem 30 in Section 3.7 in Anton (8th ed).

4) The altitude (i.e., height) of a triangle is increasing at a rate of 3.5 cm/minute while the area of the triangle is increasing at a rate of 3 square cm/minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 92 square centimeters?

The base is changing at ____________ cm/min.

5) A 16 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 2 ft/s, how fast will the foot be moving away from the wall when the top is 11 feet above the ground?

The foot will be moving at ___________ft/s.

6) A conical water tank with vertex down has a radius of 13 feet at the top and is 27 feet high. If water flows into the tank at a rate of 20 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 17 feet deep?

The depth of the water is increasing at ____________ ft/min.

7) A particle is moving along the curve y= 2 \sqrt{2 x + 6}. As the particle passes through the point (5, 8), its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

thank you

 

[soroban]

Hello, merve!

These are all standard Rate-of-Change problems.
And you can't do any of them?

\(\displaystyle \text{7) A particle is moving along the curve }y \:=\:2 \sqrt{2 x + 6}\)

\(\displaystyle \text{As the particle passes through the point }(5, 8)\text{, its }x\text{-coordinate increases at a rate of 5 units per second.}\)

\(\displaystyle \text{Find the rate of change of the distance from the particle to the origin at this instant.}\)

\(\displaystyle \text{Let the distance of the particle from the origin be }P.\)

\(\displaystyle \text{At }x = 5\text{, we want }\,\frac{dP}{dt}\,\text{ which is equal to: }\:\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\)


\(\displaystyle \text{We have: }\:y \:=\:2(2x+6)^{\frc{1}{2}}\)

\(\displaystyle \text{Differentiate with respect to time: }\;\frac{dy}{dt} \:=\:2\cdot\frac{1}{2}(2x+6)^{-\frac{1}{2}}\cdot 2 \;=\;\frac{2}{\sqrt{2x+6}}\,\frac{dx}{dt}\)

\(\displaystyle \text{At }x = 5\text{, we are told that }\frac{dx}{dt} = 5.\)

. . \(\displaystyle \text{Hence: }\:\frac{dy}{dt} \:=\:\frac{2}{\sqrt{2(5)+6}}(5) \:=\:\frac{5}{2}\)

\(\displaystyle \text{Therefore: }\:\frac{dP}{dt} \;=\;\sqrt{(5)^2 + \left(\frac{5}{2}\right)^2} \;=\;\sqrt{25 + \frac{25}{4}} \;=\;\sqrt{\frac{125}{4}} \;=\;\frac{5}{2}\sqrt{5}\,\text{ units/second}\)

 

[merve]

Thank you for your answer! But the computer doesn't accept this answer . Is it right?

I can not solve other problems, too. Can you solve them ,too? Thanks again

 

[galactus]

The same ol' related rates problems are done to death. You can find these problems, or ones very similar, if you do some googling.