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I was wondering if you would be able to check if my answers are correct:

1. An oil tank is being drained for cleaning. After t minutes there are v litres of oil left in the tank, where v(t)=40(20-t)^2, 0<=t<=20. Determine the rate of change at the time t=10.

For this question i got -400L/min by substituting 10 into the equation.

2. Consider the function f(x)=5/2+x.
a) Determine the slope at the point with x-coordinate 1

To answer this question I used the formula m=lim h->0 f(x+h)-f(x)/h. Then I got lim h->0 (5/3+h)-(5/3)/h and my answer was 1/9.

b) Determine the equation tangent to the curve f(x) at the point with x-coordinate 1

my answer for this is y=1/9x+14/9

  • Calculus - ,

    The volume is
    v(t) = 40(400-40t+t^2)

    the rate of change is the derivative, so
    v'(t) = 40(-40+2t)
    v'(10) = 40(-40+20) = -800

    2a Since your solution used 5/3, I will assume a typo in the problem.

    if you meant (5/3) + x, the slope is always 1
    if you meant 5/(3+x) then the slope at x is -5/(3+x)^2, so at x=1, the slope is -5/16

    If you used the definition of the limit, then you went wrong somewhere.

    f(x+h) = 5/(3+x+h)
    f(x) = 5/(3+x)
    f(x+h)-f(x) = [5(3+x) - 5(3+x+h)]/[(3+x)(3+x+h)]
    = -5h/[(3+x)(3+x+h)]
    divide by h and you have
    the limit as h->0 is thus

    So, at (1,5/4), the tangent line is

    y - 5/4 = -5/16 (x-1)

    see the graphs at

  • Calculus - ,

    Thanks Steve! For 2a, I meant 5/2+x, then i got 5/2+1 and then 5/3

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