Thread: Limits: Tangent and Velocity Problems

Originally Posted by asimon2008

1. A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes.

T(min)

5

10

15

25

30

V (gal)

694

444

250

28

0

(a) If P is the point (15, 250) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with t = 5,10,20,25 and 30.

Do you know the definition of "secant line" and slope? The secant line between P and Q is just the line through P and Q. With t= 5, that is (5, 694) so you are asking for the slope of the line through (15, 250) and (5, 694) and, of course, the slope of the line between and is given by . For (15, 250) and (5, 694) that is .

(b) Estimate the slope of the tangent line at P by averaging the slopes of 2 secant lines.

After you have done (a), this is easy. Choose any two of the slopes you got in (a) and average them.

(c) Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank after 15 minutes.)

Obvious first step- graph the function you are given. The put a ruler along it to draw the tangent line "by eye". Find two points on it and calculate the slope as above. When I was in high school, they gave us this method of finding a tangent line to a curve (in physics class): Place a small pocket mirror at the point on the graph where you want the tangent line. turn the mirror until the graph appears to go "smoothly" into the curve. Use the mirror, held in that position, as a straight edge to draw a line segment perpendicular to the graph. Now do the same thing to draw a line segment perpendicular to that line, giving the tangent to the curve.

3. The point P (1,1/2) lies on the curve y = x/(1+x).
(a) I f Q is the point (x,x/(1+x), use your calculator to find the slope of the secant line PQ(correct to six decimal places for the following values of x:
(i)0.5 (ii) )0.9 (iii)0.99 (iv) 0.999 (v) 1.5 (vi) 1.1

Okay, do it! As above, the slope will be where x is each of the 6 values above, in turn, and y= x/(1+x).

(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(1, 1/2)

Which of those values of x is closest to 1? What is the slope of that line?

(c) Using the slope from part (b), find an equation of the tangent line to the curve at P (1, 1/2).

You know, I presume, that a line through (1, 1/2) with slope m is of the form y= m(x-1)+ 1/2.

5. If a ball is thrown into the air with a velocity of 40ft/s, its height in feet t seconds later is given by y = 40t-16t^2.
(a) (i) 0.5 second (ii) 0.1 second (iii) 0.005 second

(b) Estimate the instantaneous velocity when t = 2.


I just need help from someone showing me how to do these ones. I have like 25 of them to do. [/QUOTE]