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1.

A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12 $ft^3$/min, how fast is the water level rising when the water is 6 inches deep?

Answer Attempt:

I'm confused at the diagram here and how to solve it. Help?

2.

A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

Answer attempt

can someone help me with this? I'm not sure of the diagram that I need to draw and how to solve it.

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  • $\begingroup$ The key to problem 1. is computing the volume of water when the height is $h.$ Can you do that? Similarly, in problem 2 you have have to be able to compute the and when $s$ feet of string have been let out. In general, in problems of this sort, the first thing to do is to get a formula for the quantity whose derivative you will have to compute, in term of the data at hand. $\endgroup$ – saulspatz Feb 22 '18 at 3:50
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    $\begingroup$ A trough would be in the shape of an upside down isosceles triangular prism $\endgroup$ – Andrew Li Feb 22 '18 at 3:55
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1.

When the height in the trough is h what is the width of the water. $$\frac{h}{b}=\frac{1}{3}\implies b=3h $$ Using this we can calculate that at height $h$ the volume is $$V=\frac12bhl=\frac32h^2l$$ Here $l$ is the length of the trough. We must now find what is $dh/dt$ given we know $dV/dt$ $$\frac{dV}{dt}=3hl\frac{dh}{dt}$$ Now put values for $dV/dt=12$, $h=6$ and $l=10$.

2.

Here we much calculate what is the angle. When kite is $x$ meters away. This makes a right triangle with height $h$ and base $x$, the angle $\theta$ is given by $$\theta=\tan ^{-1} \left(\frac{h}{x}\right)$$ We should now be able to calculate $d\theta/dt$. $$\frac{d\theta}{dt}=\frac{1}{1+(\frac{h}{x})^2}\cdot\frac{-h}{x^2}\frac{dx}{dt}$$

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  • $\begingroup$ how do you know the triangles are similar again? By the shared angle and what else? $\endgroup$ – Jwan622 Feb 22 '18 at 14:54
  • $\begingroup$ Can it be solved without converting b to h? $\endgroup$ – Jwan622 Feb 22 '18 at 15:01
  • $\begingroup$ No, at some point you will have to make that conversion doing it in the beginning is the best time $\endgroup$ – Sonal_sqrt Feb 22 '18 at 15:03
  • $\begingroup$ For 2, we just solve for x using Pythagorean theorum right? $\endgroup$ – Jwan622 Feb 22 '18 at 15:17
  • $\begingroup$ Yes that is correct $\endgroup$ – Sonal_sqrt Feb 22 '18 at 15:22

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