Depth and Normal Line

[warwick]

How do you find a line normal to a point on a particular line? I do know the derivative of the particular line at that point.

The cross section of a five-meter trough is an isosceles trapezoid with a two-meter lower base, a three-meter upper base, and altitude of 2 meters. Water is running into the trough at a rate of 2 cubic meter per minute. How fast is the water level rising when the water is 1 meter deep.

 

[royhaas]

Normal here means the two slopes are perpendicular.

 

[warwick]

Normal here means the two slopes are perpendicular.

I know what normal means. My bad. The normal line would have a slope that is the negative reciprocal of the derivative at that point.

 

[galactus]

The cross section of a five-meter trough is an isosceles trapezoid with a two-meter lower base, a three-meter upper base, and altitude of 2 meters. Water is running into the trough at a rate of 2 cubic meter per minute. How fast is the water level rising when the water is 1 meter deep.

We want \(\displaystyle \L\\\frac{dh}{dt}\) when h=1 and \(\displaystyle \L\\\frac{dV}{dt}=2\)

Let b=length of top of cross section of water.

Let h=height of water.

\(\displaystyle \L\\V=h\cdot\left(\frac{b+2}{2}\right)(5)\)

Similar triangles:

\(\displaystyle \L\\\frac{b-2}{h}=\frac{3-2}{2}=\frac{1}{2}\)

Now, solve for b in terms of h and sub into the volume equation:

\(\displaystyle \L\\b=\frac{h}{2}+2\)

\(\displaystyle \L\\V=5(\frac{\frac{h}{2}+4}{2})h=\frac{5}{4}h^{2}+10h\)

You are in terms of h alone. Differentiate:

\(\displaystyle \L\\\frac{dV}{dt}=\frac{5}{2}h\frac{dh}{dt}+10\frac{dh}{dt}\)

\(\displaystyle \L\\\frac{dh}{dt}(\frac{5}{2}(1)+10)=2\)

\(\displaystyle \L\\\frac{dh}{dt}=\fbox{\frac{2}{15}} \;\ \frac{m}{min}\)

 

[Deleted member 4993]

How do you find a line normal to a point on a particular line? I do know the derivative of the particular line at that point.
.

Let the point be (x_1,y_1)

let df/dx = m

So the slope of the normal line m_1 = -1/m

Then equation of the normal line is:

(y - y_1) = m_1 * (x - x_1)

 

[warwick]

The cross section of a five-meter trough is an isosceles trapezoid with a two-meter lower base, a three-meter upper base, and altitude of 2 meters. Water is running into the trough at a rate of 2 cubic meter per minute. How fast is the water level rising when the water is 1 meter deep.

We want \(\displaystyle \L\\\frac{dh}{dt}\) when h=1 and \(\displaystyle \L\\\frac{dV}{dt}=2\)

Let b=length of top of cross section of water.

Let h=height of water.

\(\displaystyle \L\\V=h\cdot\left(\frac{b+2}{2}\right)(5)\)

Similarity:

\(\displaystyle \L\\\frac{b-2}{h}=\frac{2}{1}\)

Now, solve for b in terms of h and sub into the volume equation.

You will be in terms of h. Differentiate, sub in the knowns and solve for dh/dt.

Similarity? Could you just work this problem out? I've moved on to the Chapter 3 review. lol. My Cal II class starts Tuesday!