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
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
-5/[(3+x)(3+x+h)]
the limit as h->0 is thus
-5/(3+x)^2
So, at (1,5/4), the tangent line is
y - 5/4 = -5/16 (x-1)
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D5%2F%283%2Bx%29%2C+y+%3D+-5%2F16+%28x-1%29%2B5%2F4
Thanks Steve! For 2a, I meant 5/2+x, then i got 5/2+1 and then 5/3
Let's check your answers:
1. An oil tank is being drained for cleaning. After t minutes, there are v liters of oil left in the tank, where v(t) = 40(20 - t)^2, 0 <= t <= 20. You want to determine the rate of change at the time t = 10.
To find the rate of change, you need to take the derivative of the function v(t) with respect to t. Let's differentiate v(t) using the power rule:
v'(t) = d/dt [40(20 - t)^2]
= 40 * 2 * (20 - t) * (-1)
= -80(20 - t)
Now, substitute t = 10 into the expression to find the rate of change at t = 10:
v'(10) = -80(20 - 10)
= -80(10)
= -800 L/min
So, the correct answer is -800 L/min, not -400 L/min.
2. Consider the function f(x) = 5/2 + x.
a) Determine the slope at the point with x-coordinate 1.
To find the slope at a specific point, you need to find the derivative of the function f(x) and evaluate it at the given x-coordinate. Let's differentiate f(x) to find the slope:
f'(x) = d/dx [5/2 + x]
= 0 + 1 (the derivative of a constant is 0)
= 1
Since the derivative is a constant, the slope at any point of f(x) is equal to this constant. So, the slope at the point with x-coordinate 1 is 1.
Therefore, your answer of 1 for the slope at x = 1 is correct.
b) Determine the equation of the tangent to the curve f(x) at the point with x-coordinate 1.
To find the equation of the tangent line, you need the slope at x = 1 and a point on the line. We already found that the slope is 1. Now, we need to find the corresponding y-coordinate at x = 1.
To find the y-coordinate, substitute x = 1 into the function f(x):
f(1) = 5/2 + 1
= 7/2
So, the point on the curve with x-coordinate 1 is (1, 7/2).
Now, we have the slope (1) and a point (1, 7/2). We can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substitute the values:
y - 7/2 = 1(x - 1)
y - 7/2 = x - 1
y = x - 1 + 7/2
y = x + 5/2
Hence, the equation of the tangent line to the curve f(x) at the point with x-coordinate 1 is y = x + 5/2.
Therefore, your answer of y = (1/9)x + 14/9 for the equation of the tangent line is incorrect. The correct answer is y = x + 5/2.