MATH 124 WS0 SOLUTION
Fall 2015
Problem 1
Alice runs along a straight line L. She is at P (1, 2) at time t = 0, and at Q(−3, 5) at time t = 5.
Find the parametric equations of L.
Solution:
Our goal here is to find equations for both x and y in terms of the independent variable t. These
could either be in slope-intercept form:
x = at + b
y = ct + d
or they could be in point-slope form:
x − x0 = a(t − t0 )
y − y0 = c(t − t0 )
Note that in the above equations x and y are variables and we can choose either point P or Q for
the values of x0 , y0 , and t0 .
In both equations, the values for a and c represent how x and y change with respect to t. To solve
for them, we can use the slope formula:
a=
x2 − x1
−3 − 1
4
=
=−
t2 − t1
5−0
5
c=
y2 − y1
5−2
3
=
=
t2 − t1
5−0
5
Now that we have a and c, the easiest thing to do is to plug them into the point-slope equations,
along with either point P or Q. It shouldn’t matter which point you choose, as you will get the
same equation if you choose to then put it into the slope-intercept form. Using the point P , where
x = 1 and y = 2 when t = 0, we get the following parametric equations:
4
4
x − 1 = − (t − 0) = − t or
5
5
3
3
y − 2 = (t − 0) = t
5
5
1
4
x=− t+1
5
3
or y = t + 2
5
Problem 2
(a) Find the equation of the tangent to the circle of radius 5 centered at the origin going through
the point P (6, 0).
Solution:
It’s usually best to begin problems like this by drawing a picture of the situation:
Here we have a line through the point P (6, 0) and another point on the circle (a, b). Since this
line is tangent to the circle, it is also perpendicular to a line through the circle’s center at (0, 0)
and the point (a, b). Before we find the equation for the line, we should try to find this point.
Note that we could also draw a line through P that intersects the circle in the fourth quadrant,
but we’ll deal with that later.
We can start by thinking of what we know about this point and the two lines. First, since the
point (a, b) is on the circle, we know that it satisfies the equation for the circle:
x2 + y 2 = 25
⇒
a2 + b2 = 25
Next, we know that since the two lines are perpendicular, their slopes are opposite reciprocals.
If we can find their slopes in terms of the variables a and b, we can then relate them using this
fact.
Slope of the line through P (0, 6):
y2 − y1
b−0
b
=
=
x2 − x1
a−6
a−6
Slope of the line through the origin (0, 0):
y2 − y1
b−0
b
=
=
x2 − x1
a−0
a
2
Since we know that these slopes are opposite reciprocals, we can write the slope of the first line
as equal to the opposite reciprocal of the slope of the second line:
b
a
=−
a−6
b
Now we have two equations and two unknowns, which we can solve via substitution:
 2
a + b2 = 25
 b = −a
a−6
b
Solving this system gives us two points, one in the first quadrant and one in the fourth:
√ !
25 5 11
,
6
6
and
√ !
25 5 11
,−
6
6
From these values we can find the slope of the lines through our two points and P (0, 6) and
write the equation for the two lines in point-slope form:
√
5 11
5
25
and y +
=√
x−
6
6
11
√
5
5 11
25
= −√
y−
x−
6
6
11
(b) Find the tangent parallel to the line y = 2x.
Solution: We can use a similar approach for this problem. First we can start with a picture:
While there are two tangent lines parallel to y = 2x, let’s first focus on the one that intersects
the circle in the fourth quadrant (the other one will intersect the circle in the second quadrant).
3
Again we can try to solve for the point of intersection, which we will call (a, b). We can use
the fact that the line through the origin and (a, b) is perpendicular to y = 2x, so its slope will
be the opposite reciprocal −1/2. Using the slope of the line and the fact that the point (a, b) is
on the circle, we get the following system of equations:

2
2

a + b = 25


b
1
=−
a
2
We can solve this to find two points of intersection:
√
√
(2 5, − 5)
and
√ √
(−2 5, 5)
Since we already know that the slope of these lines must be 2 (since they are parallel to y = 2x),
we write the equations of the tangent lines in point-slope form:
y+
√
√
5 = 2(x − 2 5)
and y −
√
√
5 = 2(x + 2 5)
Problem 3
An object is moving along a circle of radius 5 centered at O(2, 3). It starts at the point Q(2, 8). Its
angular velocity is ω = π radians per second.
(a) What are the coordinates of the object after 2 seconds?
Solution:
In two seconds the object will travel a total of 2π radians, which is one full revolution. Therefore
the object will be back at its starting position. For more difficult problems with either less
convenient velocities or time, we could always use the parametric equations for circular motion:
x(t) = x0 + r cos(ωt + α) = 2 + 5 cos(2π + π2 ) = 2
y(t) = y0 + r sin(ωt + α) = 3 + 5 sin(2π + π2 ) = 8
(b) How long does it take the object to complete a full revolution?
Solution:
As seen in part (a), it takes 2 seconds for the object to complete a full revolution.
T =
2π
2π
=
= 2 sec
ω
π
4