- #1
erik05
- 50
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Hello. Could anyone help me with a few questions that I'm stuck on? Any suggestions would be much appreciatd.
1) A man 2m tall walks away from a lampost whose light is 5m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow growing when he is 10m from the lampost?
Using pythagoras and taking the derivative I got:
[tex] \frac{dy}{dx}= \frac{-x}{y} \frac{dx}{dt}[/tex]
The derivative of x with respect to t(time) is 1.5 m/s so I think what I'm trying to find is the derivative of y with respect to t. Not too sure what to do here now but the answer is suppose to be 1m/s.
2) At 1:00 p.m. ship A was 80 km south of ship B. Ship A is sailing north at 30 km/h and ship B is sailing east at 40 km/h. How fast is the distance between them changing at 3:00 p.m.?
I think it's the times that is messing me up. Tried using the derivative of pythagoras again but where does the 80 km fit in? Answer: [tex] \frac{130}{\sqrt{17}} = 31.5 km/h [/tex]
3) A water trough is 10 m long and a cross section has the shape of an isosceles triangle that is 1m across at the top and is 50 cm high. The trough is being filled with water at a rate of 0.4 m^3/min. How fast will the water level rise when the water is 40 cm deep? Answer: 5 cm/min
Would differiantiating the equation for the area of a triangle fit into this?
Anyways, thank you in advance.
1) A man 2m tall walks away from a lampost whose light is 5m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow growing when he is 10m from the lampost?
Using pythagoras and taking the derivative I got:
[tex] \frac{dy}{dx}= \frac{-x}{y} \frac{dx}{dt}[/tex]
The derivative of x with respect to t(time) is 1.5 m/s so I think what I'm trying to find is the derivative of y with respect to t. Not too sure what to do here now but the answer is suppose to be 1m/s.
2) At 1:00 p.m. ship A was 80 km south of ship B. Ship A is sailing north at 30 km/h and ship B is sailing east at 40 km/h. How fast is the distance between them changing at 3:00 p.m.?
I think it's the times that is messing me up. Tried using the derivative of pythagoras again but where does the 80 km fit in? Answer: [tex] \frac{130}{\sqrt{17}} = 31.5 km/h [/tex]
3) A water trough is 10 m long and a cross section has the shape of an isosceles triangle that is 1m across at the top and is 50 cm high. The trough is being filled with water at a rate of 0.4 m^3/min. How fast will the water level rise when the water is 40 cm deep? Answer: 5 cm/min
Would differiantiating the equation for the area of a triangle fit into this?
Anyways, thank you in advance.