MATH 1953 Quiz 1 Solutions
March 25, 2016
1. Using any technique you like, sketch the parametrically defined graph given by x = 3t + 1, y = 6t − 2,
where t ranges from 0 to 4.
Solution. In order to graph the parametric curve define by x(t) and y(t), we will first write t in terms
of x and then substitute this t into the equation y(t). First we solve the equation x(t) for the variable
t.
x−1
(1)
x = 3t + 1 =⇒ 3t = x − 1 =⇒ t =
3
Now we must find the range of values for x. Note that since t ranges from 0 to 4, x ranges from 1 to
13 because when t = 0
x = 3(0) + 1 = 1
and when t = 4 we have
x = 3(4) + 1 = 13.
Lastly we can substitute Equation (1) into y(t), and so
x−1
y = 6t − 2 = 6
− 2 = 2(x − 1) − 2 = 2x − 4,
3
(2)
where x ranges from 1 to 13. Therefore this parametrically defined graph is given by the line y = 2x−4
where x ranges from 1 to 13.
2. (a) A curve is defined by the parametric equations x(t) = 3t2 and y(t) = 2t3 − 24t. For what value(s)
of t does the tangent line to the curve have slope 3?
Solution. Firstly we must find the derivative of this parametric curve. Recall that the derivative
of a parametric curve is given by,
dy
dy
dt
= dx
.
dx
dt
For the curve defined in this question, we have
dy
=
dx
dy
dt
dx
dt
=
6t2 − 24
4
=t−
6t
t
when t 6= 0. Now we must find for what t values does
dy
dx
(3)
= 3.
dy
4
4
= 3 =⇒ t − = 3 =⇒ t − 3 − = 0,
dx
t
t
and multiplying both sides by t gives
t2 − 3t − 4 = 0.
This last equation is quadriatic and factors as
t2 − 3t − 4 = (t − 4) (t + 1) = 0.
Therefore the t values where the tangent line to the curve has slope 3 are t = 4 and t = −1.
(b) Find an equation for the tangent line to this curve at t = 1. You do not need to simplify to
slope-intercept form.
dy
found in part (a), we can find the slope of the tangent line
Solution. Using our equation for dx
at t = 1.
dy
6(1)2 − 24
−18
=
=
= −3
dx
6(1)
6
Since t = 1 we can use the equations for x(t) and y(t) to find a point on the tangent line. This
point is (3, −22), since
x(1) = 3(1)2 = 3
and
y(1) = 2(1)3 − 24(1) = 2 − 24 = −22.
Therefore the equation of the tangent line to our parametric curve in point-slope form when t = 1
is
y + 22 = −3 (x − 3) .
(4)
This may also be written in slope-intercept form as
y = −3x − 13.
2
(5)