Page 151 - Calculus Workbook For Dummies

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Chapter 8: Using Differentiation to Solve Practical Problems

Q. Find all lines through (1, –4) either tangent 4. Get your algebra fix by finding the
3
to or normal to y = x . For each tangent equation of the tangent line that passes
line, give the point of tangency and the through (1, –4) and (2, 8).
equation of the line; for the normal lines, You can use either the point-slope form
give only the points of normalcy. or the two-point form to arrive at
y = 12 x - 16.
A. Point of tangency is (2, 8); equation of
tangent line is y = 12 x - 16. Points of nor- 5. For the normal lines, set the slope from
3
,
_
malcy are approximately (–1.539, –3.645), the general point x x i to (1, –4)
(–.335, –.038), and (.250, .016). equal to the opposite reciprocal of the
derivative and solve.
1. Find the derivative.
- 4 - x 3 - 1
y = x 3 1 - x = 3 x 2
2
5
y = l 3 x 2 - 12 x - 3 x = - 1
x
2
5
1
x
2. For the tangent lines, set the slope from 3 x + 12 x + - = 0
3
,
the general point x x i to (1, –4) x . - . 1 539 - .335 ,
,
_
equal to the derivative and solve. or .250 (Use your
- 4 - x 3 = 3 x 2 calculator .)
1 - x
3
2
- 4 - x = 3 x - 3 x 3 6. Plug these solutions into the original
function to find the points of normalcy.
3
2 x - 3 x - = 0
2
4
3
Plugging the points into y = x gives you
x = 2 _ I used my calculator. i
the three points: (–1.539, –3.645), (–.335,
3. Plug this solution into the original func- –.038), and (.250, .016).
tion to find the point of tangency.
The point is (2, 8).