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# Calculus/Rate problem solving

Question
1)A point P moves on the line y=x+5 so that dx/dt=3 units/second. Find the rate of the changing of the distance between P and Q(2,3) when P is at (-5,0).

2) the volume of a cube is increasing at a rate of 30 cm^2/sec.How fast is the total surface area of the cube increasing when the length of each edge of the cube is 10cm long?

We use V=L^3 & S=6L^2 but then what?

3) A points P is moving on the curve y=x^2-3, such that the x-coordinate is increasing at a rate of 7 cm/sec.What is the rate of change of the y-coordinate when the point P reaches (3,6)?

These problems you asked are called RELATED RATES problems.  RATES refer to the "time rate of change". In any related rates problem, there are usually two variables (say u and v).  These two variables are related (i.e. you can write an equation involving u and v), and one of the rate is given (say du/dt) and you have to find the rate of the other (i.e. dv/dt).  But the problem does not make sense (i.e. not enough information is given), unless the value of one of the quantity is given (or unless in a rare case, the relation is linear, then the value need not be given).  I like to think of "Given the rate of one variable, what is the rate of the other variable at a given moment."

1)  Let D be the distance between P(x,y) and Q(2,3).  Given the rate dx/dt=3, find dD/dt at the moment when x=-5.  If we have to related D and x, use the distance formula
D = sqrt (  (x-2)^2 + (y-3)^2  )
Since y is always x+5, D can be simplified to
D = sqrt (  (x-2)^2 + ((x+5)-3)^2  ) = sqrt ( 2(x-2)^2 )  = sqrt ( 2x^2 - 8x + 8 )

Now do an implicit time derivative (d/dt) on both sides (using chain rule) of the above equation and simplify to get:
dD/dt = 1/2 * (2x^2 - 8x + 8 )^(-1/2)  (4x-8) * dx/dt = 2(x-2)/sqrt ( 2x^2 - 8x + 8 ) * dx/dt

Now substitute the values dx/dt=3, x=-5 and one gets:
dD/dt = 2 (-5-2) / sqrt (50+40+8) = - sqrt(2) = -1.41421356
Hence the rate of the changing of the distance between P and Q(2,3) when P is at (-5,0) is about -1.14121

2)  Let S be the surface area of the cube.  Let V be the volume of the cube.  Find DS/dt given that dV/dt = 30 at the moment when L is 10 (which also mean at the moment when V = 1000).
How does one related V and S?  Well, you've already given me the two formula that V and S are dependent on L (which I assume is the length of the cube).  go ahead and square the first equation and cube the second equation, that eliminate L.  Oh, by the way, the unit for rate of increase of volume of the cube should be in  cm^3/sec (we read that as cubic centimeter per second)  not cm^2/sec.
So 216 V^2 = S^3.  (which means at the moment V = 1000, or L = 10, S = 600)
Do an implicit time derivative (d/dt) on both sides (using chain rule) of the above equation:
216 (2 V) * dV/dt   = (3 S) * dS/dt

Substitute V = 1000 (which means S = 600) and dV/dt = 30 into the above equation, one gets:
216 (2000)* (30)   = (1800) * dS/dt

now, can I leave it to you to find the rate of change of the total surface area?

3)  Given dx/dt = 7.  find dy/dt when x=3 and y = 6.

y = x^2 - 3.
Do an implicit time derivative (d/dt) on both sides (using chain rule) of the above equation:
dy/dt = 2*x* dx/dt

Subs x = 3, dx/dt = 7 and woalah!  one gets dy/dt !

Calculus

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#### Amos

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I can answer all calculus question. I am a Math Lecturer and I teach Math in a College, usually Calculus 1, Calculus 2 and Calculus 3 and Linear Algebra.

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I have been teaching in this college for more than 12 years. Prior to that, I taught for three years as Visiting Assistant Professor.

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I got my doctorate from Univ. of Rochester in Algebraic Toppology. I got a MA from Univ of Rochester, an MA from York Univ. in Toronto and almost did my M.Sc in the National Univ. of Singapore.