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.
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 $d\theta/dt$. $$\frac{d\theta}{dt}=\frac{1}{1+(\frac{h}{x})^2}\cdot\frac{-h}{x^2}\frac{dx}{dt}$$