10.2 Calculus with Parametric Curves by SGLee - HSKim - JHLee
1-2. Find 
.
1. 
.
Calculus-Sec-10-2-Solution
10.2 Calculus with Parametric Curves by SGLee - HSKim - JHLee
1-2. Find .
1.
.
2. 
.
3-6. Find an equation of the tangent to the curve at the point.
3. 
. 
At the point with parameter value 
, the slope is 
.
Slope of the tangent at 
is 
. Hence, the equation of the tangent line is 
. <-- y=-2+3/5(x+2)=3/5x-4/5
4. 
At the point with parameter value 
, the slope is 
. <-- 역수로 고쳐야 함
Slope of the tangent at 
is 
.
Hence, the equation of the tangent line is 
. <-- y=sqrt(3)/2 x - 1/2
5. 
At the point with parameter value 
, the slope is 
.
The point (1,1) corresponds to the parameter value 
,
so the slope of the tangent at the point is 
.
Hence, the equation of the tangent line is 
.
6. 
http://matrix.skku.ac.kr/cal-lab/cal-9-2-6.html
At the point with parameter 
, the slope is 
.
If 
. It’s slope of the tangent at 
is 
.
Hence, the equation of the tangent line is 
.
7-10. Find 
. For what values of 
is the curve concave upward or downward?
7. 
http://matrix.skku.ac.kr/cal-lab/cal-9-2-7.html
,
.
⇒ Since 
, 
for all 
.
Hence, the curve is downward everywhere.
8. 
,
.
⇒ If 
, 
. Thus if 
, 
.
Hence, the curve is concave upward on 
,
and is concave downward on 
.
9. 
, 
, 
http://matrix.skku.ac.kr/cal-lab/cal-9-2-9.html
.
If 
, 
. Thus if 
, 
.
Hence, the curve is upward on 
, and the curve is downward on 
.
Note that this curve is an ellipse.
10. 
11-12. Find the point on the curve where the tangent is horizontal or vertical.
11. 
http://matrix.skku.ac.kr/cal-lab/cal-9-2-11.html
so 
so 
The curve has horizontal tangents at 
,
and vertical tangents at 
.
12. 
, so 
for 
.
Points at which the tangent is horizontal and 
,
so 
for 
.
The curve has horizontal tangents at 
,
and vertical tangents at 
.
13. Show that the curve 
, 
has two tangents at (0, 0) and find their equations.
.
Now 
is 0 when 
, so there are two tangents at the points (0,0)
since both 
correspond to the origin.
The tangent corresponding to 
has slope 2, and its equation is 
.
The tangent corresponding to 
has slope -2, and its equation is 
.
14. At what point does the curve 
, 
cross itself?
Find the equations of both tangents at that point.
http://matrix.skku.ac.kr/cal-lab/cal-9-2-14.html
From the figure it is clear that at (0, 0) the curve does cross itself.
Moreover, 
is 0 when 
, 
, 
, 
.
The tangent corresponding to t has slope 1 or -1.
Then the equations of tangents at (0,0) are 
.
15. At what points on the curve 
is the tangent parallel to the line with equations 
?
Given latter function’s slope is 
,
so we should find 
when the former function’s slope is 
.
, 
. We can get 
is 
or 
.
When 
is 
or 
, the former function’s tangent parallel to the latter’s.
16. Find the area bounded by the curve 
, 
and the line 
.
When 
or 
, 
.
When 
, x=-3/2. When 
, x=3/2 .
∵
17. Find the area bounded by the curve 
, 
, and the lines 
and 
.
http://matrix.skku.ac.kr/cal-lab/cal-9-2-17.html
The curve 
intersects the 
-axis when 
.
Then, the corresponding values of 
are 
.
18. Find the area of a region enclosed by the astroid 
.
http://matrix.skku.ac.kr/cal-lab/cal-9-2-18.html
∴(
)
19. Find the arc length of the curve defined by
.
. Then, 
.
Hence, 
.
20. Find the arc length of the curve defined by
.
, 
=sqrt(2)*e^pi - sqrt(2)
21. Find the arc length of the curve defined by
, 
; 
.
http://matrix.skku.ac.kr/cal-lab/cal-9-2-21.html
22. Find the arc length of the curve defined by
, 
; 
.
http://matrix.skku.ac.kr/cal-lab/cal-9-2-22.html
23. Find the length of one arch of the cycloid 
, 
.
. Then, 
.
Hence, 
.
24-26. Find the area of the surface obtained by rotating the given curve about the 
-axis.
24. 
, 
: 
. Then, 
.
Hence, 
.
25. 
, 
: 
. Then, 
.
Hence,
.
26. 
, 
; 
27-29. Find the area of the surface generated by rotating the given curve about 
-axis.
27. 
, 
; 
http://matrix.skku.ac.kr/cal-lab/cal-9-2-27.html
, 
= 2/1215*(247*sqrt(13) + 64)*pi
28. 
, 
; 
29. 
, 
; 
