Rate of Change

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Examples : Activity 1

1. Find the rate of change of the area of a circle with respect to its radius r when
   a) r = 3 cm
   b) r= 4 cm
   

   Solution 👉 Click Here

   The area of circle is
   \( A=\pi r^2\)
   The rate of change of the area of a circle with respect to its radius r is
   \( \frac{dA}{dr}=2 \pi r\)
   So, the rate of change of the area of a circle with respect to its radius r when
   a) r = 3 cm is \( \frac{dA}{dr}=2 \pi r=6 \pi\)
   b) r= 4 cm is \( \frac{dA}{dr}=2 \pi r=8 \pi\)
   
2. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
   

   Solution 👉 Click Here

   Suppose that
   Length of radius=r=10
   Area of circle= A
   Given that
   The radius of a circle is increasing uniformly at the rate of 3 cm/s
   \(\frac{dr}{dt}=3\)
   Now, the rate at which the area of the circle is increasing is
   \(\frac{dA}{dt}=\frac{dA}{dr}. \frac{dr}{dt}\)
   or \(\frac{dA}{dt}=\frac{d \pi r^2}{dr}. 3\)
   or \(\frac{dA}{dt}= 2\pi r. 3\)
   or \(\frac{dA}{dt}=6\pi r\)
   or \(\frac{dA}{dt}= 6\pi .10\)
   or \(\frac{dA}{dt}=60\pi\)
   
3. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
   

   Solution 👉 Click Here

   Suppose that
   The edge of the variable cube =l=10
   The volume of the variable cube =V
   Given that
   The edge of a variable cube is increasing at the rate of 3 cm/s
   \( \frac{dl}{dt}=3\)
   The rate of change in the volume of the cube is
   \( \frac{dV}{dt}= \frac{dV}{dl}. \frac{dl}{dt}\)
   or\( \frac{dV}{dt}= 3l^2. \frac{dl}{dt}\)
   or\( \frac{dV}{dt}= 3.10^2. 3\)
   or\( \frac{dV}{dt}= 900 cm^3/s\)
   
4. The radius of a circle is increasing at the rate of 7 cm/s. What is the rate of increase of its circumference?
   

   Solution 👉 Click Here

   Suppose that
   The radius of a circle =r
   The circumference of a circle =C
   Given that
   The radius of a circle is increasing at the rate of 0.7 cm/s
   \( \frac{dr}{dt}=7\)
   The rate of change in circumference of a circle is
   \( \frac{dC}{dt}= \frac{dC}{dr}. \frac{dr}{dt}\)
   or\( \frac{dC}{dt}= 2 \pi. \frac{dr}{dt}\)
   or\( \frac{dC}{dt}= 2 \pi. 7\)
   or\( \frac{dC}{dt}= 14 \pi. cm/s\)
   
5. A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
   

   Solution 👉 Click Here

   Suppose that
   The radius of a sphere =r
   The volume of a sphere =V
   The rate of change in volume with the radius of a sphere is
   \( \frac{dV}{dr}= \frac{d (\frac{4}{3}\pi r^3)}{dr} \)
   or\( \frac{dV}{dr}= 4 \pi r^2 \)
   
6. A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}(2x+1)\). Find the rate of change of its volume with respect to x.
   

   Solution 👉 Click Here

   Suppose that
   The diameter of a sphere =\(\frac{3}{2}(2x+1)\)
   So, radius of a sphere =\( \frac{3}{4}(2x+1) \)
   The volume of a sphere =V
   The rate of change in volume with the radius of a sphere with respect to x is
   \( \frac{dV}{dx}= \frac{d (\frac{4}{3}\pi r^3)}{dx} \)
   or\( \frac{dV}{dx}= \frac{d (\frac{4}{3}\pi \left ( (\frac{3}{4}(2x+1) \right)^3}{dx} \)
   or\( \frac{dV}{dx}= \frac{9}{16} \pi \frac{d (2x+1) ^3}{dx} \)
   or\( \frac{dV}{dx}= \frac{9}{16} \pi 3(2x+1)^2 .2 \)
   or\( \frac{dV}{dx}= \frac{9}{16} \pi 6(2x+1)^2 \)
   or\( \frac{dV}{dx}= \frac{27}{8} \pi (2x+1)^2 \)
   
7. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
   

   Solution 👉 Click Here

   Suppose that
   The radius of waves move in circles =r=8
   The area of waves move in circles =A
   Given that
   The radius of waves move in circles is at the speed of 5 cm/s
   \( \frac{dr}{dt}=5\)
   The rate of change in area of waves move in circles is
   \( \frac{dA}{dt}= \frac{dA}{dr}. \frac{dr}{dt}\)
   or\( \frac{dA}{dt}= 2 \pi r. \frac{dr}{dt}\)
   or\( \frac{dA}{dt}= 2 \pi r. 5 \)
   or\( \frac{dA}{dt}= 10 \pi r \)
   or\( \frac{dA}{dt}= 10 \pi .8 \)
   or\( \frac{dA}{dt}= 40 \pi \)
   
8. Air is being pumped into a spherical balloon such that its radius increases at a rate of 0.75 in/min. Find the rate of change of its volume when the radius is 5 inches.

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Examples : Activity 2

1. A particle moves along the curve \(6y = x^3 +2\). Find the points on the curve at which the y-coordinate changes 8 times as fast as the x-coordinate.
   

   Solution 👉 Click Here

   Suppose that
   The coordinate on the curve is =(x,y)
   Given that
   y-coordinate changes 8 times as fast as the x-coordinate
   \( \frac{dy}{dt}=8 \frac{dx}{dt}\)
   Now
   \(6y = x^3 +2\)
   or\(6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}\)
   or\( \frac{dy}{dt} = \frac{x^2}{2} \frac{dx}{dt}\)
   or\( 8 \frac{dx}{dt} = \frac{x^2}{2} \frac{dx}{dt}\)
   or\( 16 = x^2\)
   or\( x= \pm 4\)
   Now,
   if \(x=4\) then \(y=\)
   if \(x=-4\) then \(y=\)
   
2. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?
   

   Solution 👉 Click Here

   Suppose that
   The distance of foot of the ladder from the wall =x=4
   The height of ladder on the wall =y =(3)
   The length of ladder =5
   So,
   \(x^2+y^2=5^2\)
   Given that
   The bottom of the ladder is pulled at the rate of 2cm/s.
   \( \frac{dx}{dt}=2 \)
   Now
   The rate of change in decreasing the height on the wall is
   \( \frac{dy}{dt}=? \)
   Here
   \(x^2+y^2=5^2\)
   or \(2x \frac{dx}{dt} +2y \frac{dy}{dt}=0\)
   or \(2x .2 +2y \frac{dy}{dt}=0\)
   or \(2x + y \frac{dy}{dt}=0\)
   or \(\frac{dy}{dt}=- \frac{2x}{y}\)
   or \(\frac{dy}{dt}=- \frac{2.4}{3}\)
   or \(\frac{dy}{dt}=- \frac{8}{3} cm/s\)
   
3. The volume of a cube is increasing at the rate of \(8 cm^3/s\). How fast is the surface area increasing when the length of an edge is 12 cm?
   

   Solution 👉 Click Here

   Suppose that
   Length of cube=s=12
   Volume of cube= V
   Area of cube=A
   Given that
   The volume of a cube is increasing at the rate of \(8 cm^3/s\)
   \(\frac{dV}{dt}=8\)
   or \(\frac{dV}{ds}. \frac{ds}{dt}=8\)
   or \(\frac{ds^3}{ds}. \frac{ds}{dt}=8\)
   or \(3s^2 \frac{ds}{dt}=8\)
   or \( \frac{ds}{dt}=\frac{8}{3s^2}\)
   Now, change in the surface area is
   \(\frac{dA}{dt}=frac{dA}{ds}. \frac{ds}{dt}\)
   or \(\frac{dA}{dt}=\frac{d6s^2}{ds}. \frac{ds}{dt}\)
   or \(\frac{dA}{dt}=12s .\frac{8}{3s^2}\)
   or \(\frac{dA}{dt}=\frac{32}{s}\)
   or \(\frac{dA}{dt}=\frac{32}{12}\)
   or \(\frac{dA}{dt}=\frac{8}{3} cm^2/s\)
   
4. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y= 6cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
   

   Solution 👉 Click Here

   Given that
   Length of rectangle=x=8
   Breadth of rectangle=y=6
   Now,
   The length x of a rectangle is decreasing at the rate of 5 cm/minute.
   \(\frac{dx}{dt}=-5\)
   The width y is increasing at the rate of 4 cm/minute.
   \(\frac{dy}{dt}=4\)
   The rates of change of (a) the perimeter of rectangle is
   \(\frac{dP}{dt}=frac{d2(x+y)}{dt}\)
   or \(\frac{dP}{dt}=2(\frac{dx}{dt}+\frac{dy}{dt}) \)
   or \(\frac{dP}{dt}=2(-5+4) \)
   or \(\frac{dP}{dt}=-2 \)
   Next, the rates of change of (b) the area of the rectangle is
   \(\frac{dA}{dt}=\frac{d (xy)}{dt}\)
   or \(\frac{dA}{dt}=y \frac{dx}{dt}+x \frac{dy}{dt}) \)
   or \(\frac{dA}{dt}=6 (-5)+4 (8) \)
   or \(\frac{dA}{dt}=2 \)
   
5. Sand is pouring from a pipe at the rate of 12 cm^3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
   

   Solution 👉 Click Here

   Suppose that
   Volume of cone=V
   Height of cone= h =4
   Radius of cone=r (r=6h)
   Given that
   The rate of change in volume of a cone is pouring sand from a pipe at the rate of 12 cm^3/s
   \(\frac{dV}{dt}=12\)
   Now,
   The change in the height of the cone is
   \(\frac{dh}{dt}=?\)
   Here
   \(\frac{dV}{dt}=12\)
   or\( \frac{dV}{dh} \frac{dh}{dt}=12\)
   or\( \frac{d \left ( \frac{1}{3} \pi r^2h \right ) }{dh} \frac{dh}{dt}=12\)
   or\( \frac{d \left ( \frac{1}{3} \pi (6h)^2h \right ) }{dh} \frac{dh}{dt}=12\)
   or\( \frac{d \left ( \frac{1}{3} \pi 36h^3 \right ) }{dh} \frac{dh}{dt}=12\)
   or\( \frac{d \left ( 12 \pi h^3 \right ) }{dh} \frac{dh}{dt}=12\)
   or\( 36 \pi h^2 \frac{dh}{dt}=12\)
   or\( \frac{dh}{dt}=\frac{12}{36 \pi h^2}\)
   or\( \frac{dh}{dt}=\frac{12}{36. \pi. 4^2}\)
   or\( \frac{dh}{dt}=\frac{1}{3. \pi /4^2}\)
   or\( \frac{dh}{dt}=\frac{1}{48 \pi }\)
   
6. The position of a particle on a line is given by \(s(t) = t 3 − 3 t 2 − 6 t + 5\) , where t is measured in seconds and s is measured in feet. Find
   a. The velocity of the particle at the end of 2 seconds.
   b. The acceleration of the particle at the end of 2 seconds.