Using Implicit Differentiation to Solve Related Rates Problems

Key Questions

• If #x^2+y^2=25# and #dy/dt=6#, how do you find #dx/dt# when #y=4# ?

  There are two values depending on the point.

  #{dx}/{dt}={(-8 " at " (3,4)),(8 " at "(-3,4)):}#

  Let us look at some details.

  First, let us find the values of #x#.

  By plugging in #y=4# into #x^2+y^2=25#,

  #x^2+16=25 Rightarrow x^2=9 Rightarrow x=pm3#

  Now, let us find some derivatives.

  By differentiating with respect to #t#,

  #d/{dt}(x^2+y^2)=d/{dt}(25) Rightarrow 2x{dx}/{dt}+2y{dy}/{dt}=0#

  by dividing by #2x#,

  #Rightarrow {dx}/{dx}+y/x{dy}/{dt}=0#

  by subtracting #y/x{dy}/{dt}#,

  #Rightarrow {dx}/{dt}=-y/x{dy}/{dt}#

  Since #y=4#, #x=pm3#, and #{dy}/{dt}=6#,

  #{dx}/{dt}=-{4}/{pm3}(6)=pm8#

  Wataru · 9 · Sep 21 2014
• If the radius of a sphere is increasing at a rate of 4 cm per second, how fast is the volume increasing when the diameter is 80 cm?

  Answer:

  12,800cm3s

  Explanation:

  This is a classic Related Rates problems. The idea behind Related Rates is that you have a geometric model that doesn't change, even as the numbers do change.

  For example, this shape will remain a sphere even as it changes size. The relationship between a where's volume and it's radius is

  #V=4/3pir^3#

  As long as this geometric relationship doesn't change as the sphere grows, then we can derive this relationship implicitly, and find a new relationship between the rates of change.

  Implicit differentiation is where we derive every variable in the formula, and in this case, we derive the formula with respect to time.

  So we take the derivative of our sphere:

  #V=4/3pir^3#

  #(dV)/(dt)=4/3pi(3r^2)(dr)/dt#

  #(dV)/(dt)=4pir^2(dr)/dt#

  We were actually given #(dr)/(dt)#. It's #4(cm)/s#.

  We are interested in the moment when the diameter is 80 cm, which is when the radius will be 40 cm.

  The rate of increase of the volume is #(dV)/(dt)#, which is what we are looking for, so:

  #(dV)/(dt)=4pir^2(dr)/dt#

  #(dV)/(dt)=4pi(40cm)^2(4(cm)/s)#

  #(dV)/(dt)=4pi(1600cm^2)(4(cm)/s)#

  #(dV)/(dt)=4pi(1600cm^2)(4(cm)/s)#

  #(dV)/(dt)=12,800(cm^3)/s#

  And the units even work out correctly, since we should get a volume divided by time.

  Hope this helps.

  

  Paul Belliveau · 24 · Oct 6 2014

• If a cylindrical tank with radius 5 meters is being filled with water at a rate of 3 cubic meters per minute, how fast is the height of the water increasing?
• If the radius of a sphere is increasing at a rate of 4 cm per second, how fast is the volume increasing when the diameter is 80 cm?
• If #y=x^3+2x# and #dx/dt=5#, how do you find #dy/dt# when #x=2# ?
• If #x^2+y^2=25# and #dy/dt=6#, how do you find #dx/dt# when #y=4# ?
• How do you find the rate at which water is pumped into an inverted conical tank that has a height of 6m and a diameter of 4m if water is leaking out at the rate of #10,000(cm)^3/min# and the water level is rising #20 (cm)/min#?
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• At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 5 square meters and the radius is increasing at the rate of 1/3 meters per minute?
• At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?
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