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Applications of Derivatives Test - 4
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Applications of Derivatives Test - 4
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  • Question 1/10
    1 / -0.25

    The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.​

  • Question 2/10
    1 / -0.25

    The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x2 + 22x + 35. Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant

    Solutions

  • Question 3/10
    1 / -0.25

    Find the approximate change in total surface area of a cube of side x metre caused by increase in side by 1%.​

  • Question 4/10
    1 / -0.25

    The volume of cube is increasing at the constant rate of  3 cm3 /s. Find the rate of change of edge of the cube when its edge is 5 cm.​

    Solutions

    Let V be the instantaneous volume of the cube.
    dV/dt = 3 cm3 /s
    Let x be the side of the cube.
    V = x3
    dV/dt = 3x2 * (dx/dt)
    3 = 3*(52 )*(dx/dt)
    So, dx/dt = 1/25 cm/s

  • Question 5/10
    1 / -0.25

    The total cost associated with the production of x units of a product is given by C(x) = 5x2 + 14x + 6. Find marginal cost when 5 units are produced

    Solutions

    C(x) = 5x2 + 14x + 6
    Marginal Cost M(x) = C ’(x)
    M(x) = 10x +14
    So, M(5) = 50 + 14
    = 64 Rs

  • Question 6/10
    1 / -0.25

    At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.

    Solutions

    Tan(z) = h x

  • Question 7/10
    1 / -0.25

    A stone is dropped into a quiet lake and waves move in circles at a speed of 2cm per second. At the instant, when the radius of the circular wave is 12 cm, how fast is the enclosed area changing ?​

    Solutions

    Rate of increase of radius dr/dt = 2 cm/s
    Area of circle A = πr2
    dA/dt = π*(2r)*(dr/dt)
    = π*(24)*2
    = 48 πcm2 /s
    Rate of increase of area is 48 πcm2 /s (increasing as it is positive).

  • Question 8/10
    1 / -0.25

    Using approximation find the value of  

    Solutions

    Let x=4, Δx=0.01

    y=x^½= 2

    y+Δy = (x+ Δx)^½= (4.01)^½

    Δy = (dy/dx) * Δx

    Δy = (x^(-1/2))/2 * Δx

    Δy = (½)*(½) * 0.01

    Δy = 0.25 * 0.01

    Δy = 0.0025

    So, (4.01)^½= 2 + 0.0025 = 2.0025

  • Question 9/10
    1 / -0.25

    Find the approximate value of  f(10.01) where f(x) = 5x2 +6x + 3 ​

    Solutions

    f(x) = 5x2 +6x + 3
    f(10.01) = 5*(10.01)2 + 6*(10.01) + 3
    To find (10.01)2
    Let p=10, Δp=0.01
    y=p2 = 100
    y+Δy = (p+ Δp)2 = (10.01)2
    Δy = (dy/dp) * Δp
    Δy = 2*p* Δx
    Δy = 2*10* 0.01
    Δy = 20 * 0.01
    Δy = 0.2
    So, (10.01)2 = y + Δy
    = 100.2
    So,
    f(10.01) = 5*(100.2) + 6*(10.01) + 3
    = 501 + 60.06 + 3
    = 564.06

  • Question 10/10
    1 / -0.25

    Given a function y = f(x) . Let Δx be the very small change in the value of x , then the corresponding change in the value of y that is Δy is approximately given by

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