Answer to Question #175733 in Calculus for Shahir Sheikh

Question

Let C be the curve defined by the parametric equations:

x = 4t² + 4t

y = 2cos(t) + 3

t ∈ [0,π]

a) Find the value of t on the indicated domain when the x-coordinate of the curve is 3. Write your answer as a fraction

t =

b) Find the value of t on the indicated domain when the y-coordinate of the curve is 2. Round your answer to 3 decimal places 

t =

c) Find the y-intercept of the curve.

The y-intercept is at ( 0,_)

Answer 1

а) The x-coordinate of the curve is 3, if:

$$4t^2+4t=3$$ $$4t^2+4t-3=0$$ $$D=4^2-4\cdot4\cdot(-3)=16+48=64$$ $$t_1=\frac{-4-\sqrt{64}}{2\cdot4}=\frac{-4-8}{8}=-1.5$$ $$t_2=\frac{-4+\sqrt{64}}{2\cdot4}=\frac{-4+8}{8}=0.5$$

Since $t_1 \notin [0; \pi]$, then x =3 at t=0.5

b) The y-coordinate of the curve is 2, if:

$$2cos(t) + 3=2$$ $$2cos(t) =-1$$ $$cos(t) = -\frac{1}{2}$$

This equation has one solution on the interval $[0; \pi]$: $t = \frac{2\pi}{3}$

c) The y-intercept is an (x,y) point with x=0. Hence,

$$4t^2+4t=0$$ $$4t(t+1)=0$$ $$\left[ \begin{gathered} t=0\\t+1=0 \end{gathered} \right.$$ $$\left[ \begin{gathered} t=0\\t=-1 \end{gathered} \right.$$

Since $t=-1 \notin [0; \pi]$, then x=0 at t=0. Find y at t=0:

$$y=2cos(0)+3=2+3=5$$

Hence, the  y-intercept is at (0,5)