Solutions to Problem Set 31

Math 211-03

11-21-2017

[Parametric equations]

1. Two lines are given in parametric form:

$$x = s + 1, \quad y = 3 s + 7 \quad\hbox{and}\quad x = -5 t + 4, \quad y = 2 t + 3.$$

Find the point where the lines intersect.

Set the two x-expressions equal:

$$\eqalign{ s + 1 & = -5 t + 4 \cr s & = -5 t + 3 \cr}$$

Equate the expressions for y, then substitute $s = -5 t + 3$ :

$$\eqalign{ 3 s + 7 & = 2 t + 3 \cr 3 (-5 t + 3) + 7 & = 2 t + 3 \cr -15 t + 16 & = 2 t + 3 \cr -17 t & = -13 \cr \noalign{\vskip2pt} t & = \dfrac{13}{17} \cr}$$

Substitute this into $x = -5 t +
   4$ and $y = 2 t + 3$ to get $x =
   \dfrac{3}{17}$ and $y = \dfrac{77}{17}$ .

The intersection point is $\left(\dfrac{3}{17}, \dfrac{77}{17}\right)$ .


2. Find the equation of the tangent line to

$$x = (t^2 + 1)^2, \quad y = t^3 + 2 t + 1, \quad\hbox{at}\quad t = 1.$$

When $t = 1$ , I have $x =
   4$ and $y = 4$ . The point of tangency is $(4, 4)$ .

$$\der y x = \dfrac{\der y t}{\der x t} = \dfrac{3 t^2 + 2}{4 t(t^2 + 1)}.$$

Setting $t = 1$ gives $\der y x = \dfrac{5}{8}$ . The tangent line is

$$y - 4 = \dfrac{5}{8}(x - 4), \quad\hbox{or}\quad y = \dfrac{5}{8} x + \dfrac{3}{2}.\quad\halmos$$


3. Find the equation of the tangent line to

$$x = t + 3 e^{2 t}, \quad y = -4 t + 5 e^{2 t}, \quad\hbox{at}\quad t = 0.$$

When $t = 0$ , I have $x =
   3$ and $y = 5$ .

$$\der y x = \dfrac{\der y t}{\der x t} = \dfrac{-4 + 10 e^{2 t}}{1 + 6 e^{2 t}}.$$

Setting $t = 0$ gives $\der y x = \dfrac{6}{7}$ . The tangent line is

$$y - 5 = \dfrac{6}{7} (x - 3), \quad\hbox{or}\quad y = \dfrac{6}{7} x + \dfrac{17}{7}.\quad\halmos$$


4. Consider the parametric curve

$$x = t^3 - 3 t + 1, \quad y = t^2 - 4 t + 5.$$

(a) Find the value(s) of t for which the curve has a horizontal tangent line.

(b) Find the value(s) of t for which the curve has a vertical tangent line.

I have

$$\der y x = \dfrac{\der y t}{\der x t} = \dfrac{2 t - 4}{3 t^2 - 3} = \dfrac{2(t - 2)}{3(t + 1)(t - 1)}.$$

(a) The tangent line is horizontal when $\der y x = 0$ , which occurs when $t = 2$ .

(b) The tangent line is vertical when $\der y x$ is undefined, which occurs when $t = \pm 1$ .


5. Find $\dfrac{d^2y}{dx^2}$ for the curve

$$x = t^2 + 1, \quad y = t^2 + 3 t + 1, \quad\hbox{at}\quad t = 1.$$

$$\der y x = \dfrac{2 t + 3}{2 t},$$

$$\dfrac{d^2y}{dx^2} = \dfrac{\der {} t \left(\der y x\right)}{\der x t} = \dfrac{\dfrac{(2 t)(2) - (2 t + 3)(2)}{(2 t)^2}}{2 t} = \dfrac{-6}{8 t^3} = \dfrac{-3}{4 t^3}.$$

When $t = 1$ , I have $\dfrac{d^2y}{dx^2} =\dfrac{-3}{4}$ .


6. Find $\dfrac{d^2y}{dx^2}$ for the curve

$$x = t^3, \quad y = (t + 1)^2, \quad\hbox{at}\quad t = 1.$$

$$\der y x = \dfrac{2(t + 1)}{3 t^2},$$

$$\dfrac{d^2y}{dx^2} = \dfrac{\der {} t \left(\der y x\right)}{\der x t} = \dfrac{\dfrac{(3 t^2)(2) - 2(t + 1)(6 t)}{(3 t^2)^2}}{3 t^2} = \dfrac{-12 t - 6 t^2}{27 t^6} = \dfrac{-4 - 2 t}{9 t^5}.$$

When $t = 1$ , I have $\dfrac{d^2y}{dx^2} = -\dfrac{6}{9} = -\dfrac{2}{3}$ .


7. Find $\dfrac{d^2y}{dx^2}$ for the curve

$$x = t^2 + 3 t + 1, \quad y = t^3 + 1, \quad\hbox{at}\quad t = 1.$$

$$\der y x = \dfrac{3 t^2}{2 t + 3},$$

$$\dfrac{d^2y}{dx^2} = \dfrac{\der {} t \left(\der y x\right)}{\der x t} = \dfrac{\dfrac{(2 t + 3)(6 t) - (3 t^2)(2)}{(2 t + 3)^2}}{2 t + 3} = \dfrac{(2 t + 3)(6 t) - (3 t^2)(2)}{(2 t + 3)^3}.$$

When $t = 1$ , I have

$$\dfrac{d^2y}{dx^2} = \dfrac{(5)(6) - (3)(2)}{5^3} = \dfrac{24}{125}.\quad\halmos$$


8. You can think of a parametric curve $(x(t), y(t))$ as being traced out by a particle moving in the x-y-plane.

$$\hbox{\epsfysize=1.25in \epsffile{parametric-equations1.eps}}$$

The velocity vector of the curve is given by $\left(\der x t, \der y
   t\right)$ . It points in the direction of the particle's motion.

$$\hbox{\epsfysize=1.25in \epsffile{parametric-equations2.eps}}$$

The speed is the length of the velocity vector, and is given by $\sqrt{\left(\der x t\right)^2 + \left(\der y t\right)^2}$ .

Find the velocity vector and the speed for

$$x = t^2 + t + 1, \quad y = -t^4 + 7, \quad\hbox{at}\quad t = 1.$$

$$\left(\der x t, \der y t\right) = \left(2 t + 1, -4 t^3\right).$$

When $t = 1$ , the velocity vector is $(3, -4)$ .

At this point, the speed is $\sqrt{3^2 + (-4)^2} = 5$ .


One hears only those questions for which one is able to find answers. - Friedrich Nietzsche


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