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Water draining from a funnel

Water is draining from the bottom of a cone-shaped funnel at the rate of 0.0 3 ft 3 /sec . The height of the funnel is 2 ft and the radius at the top of the funnel is 1 ft . At what rate is the height of the water in the funnel changing when the height of the water is 1 2 ft ?

Step 1: Draw a picture introducing the variables.

A funnel is shown with height 2 and radius 1 at its top. The funnel has water to height h, at which point the radius is r.
Water is draining from a funnel of height 2 ft and radius 1 ft. The height of the water and the radius of water are changing over time. We denote these quantities with the variables h and r , respectively.

Let h denote the height of the water in the funnel, r denote the radius of the water at its surface, and V denote the volume of the water.

Step 2: We need to determine d h d t when h = 1 2 ft . We know that d V d t = −0.03 ft/sec .

Step 3: The volume of water in the cone is

V = 1 3 π r 2 h .

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, r h = 1 2 or r = h 2 . Using this fact, the equation for volume can be simplified to

V = 1 3 π ( h 2 ) 2 h = π 12 h 3 .

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time t , we obtain

d V d t = π 4 h 2 d h d t .

Step 5: We want to find d h d t when h = 1 2 ft . Since water is leaving at the rate of 0.0 3 ft 3 /sec , we know that d V d t = −0.03 ft 3 /sec . Therefore,

−0.03 = π 4 ( 1 2 ) 2 d h d t ,

which implies

−0.03 = π 16 d h d t .

It follows that

d h d t = 0.48 π = −0.153 ft/sec .
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At what rate is the height of the water changing when the height of the water is 1 4 ft ?

−0.61 ft/sec

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Key concepts

  • To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time.
  • In terms of the quantities, state the information given and the rate to be found.
  • Find an equation relating the quantities.
  • Use differentiation, applying the chain rule as necessary, to find an equation that relates the rates.
  • Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.

For the following exercises, find the quantities for the given equation.

Find d y d t at x = 1 and y = x 2 + 3 if d x d t = 4 .

8

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Find d x d t at x = −2 and y = 2 x 2 + 1 if d y d t = −1 .

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Find d z d t at ( x , y ) = ( 1 , 3 ) and z 2 = x 2 + y 2 if d x d t = 4 and d y d t = 3 .

13 10

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For the following exercises, sketch the situation if necessary and used related rates to solve for the quantities.

[T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, Ω ) is given by the equation 1 R = 1 R 1 + 1 R 2 . If R 1 is increasing at a rate of 0.5 Ω / min and R 2 decreases at a rate of 1.1 Ω/min , at what rate does the total resistance change when R 1 = 20 Ω and R 2 = 50 Ω / min ?

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A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

A right triangle is formed by a ladder leaning up against a brick wall. The ladder forms the hypotenuse and is 10 ft long.

2 3 ft/sec

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A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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