Question Type 8: Finding the time of change in both directions for a given object Exercises

Find the local maximum of $x(t)$ by determining the time(s) at which $x(t)$ attains an extremum and identifying which is a maximum, given $x(t)=3+t(9-t^2).$

Show that the object never comes to rest (i.e. its speed never becomes zero) for: $x(t) = 3 + t(9 - t^2), \quad y(t) = 4 + t(11 + t)$

For the parametric trajectory given by

find the times $t$ when the object changes direction in the $x$-direction.

Determine the interval(s) of $t$ for which the object moves to the right (i.e., $\frac{dx}{dt} > 0$), given that

Find the time at which the $x$-coordinate equals 10, given $x(t)=3+9t-t^3.$

Determine whether there is any time $t$ such that the tangent to the trajectory $x(t)=3+t(9-t^2),\quad y(t)=4+t(11+t)$ has slope $1$, i.e. $\frac{dy}{dx}=1.$

For the trajectory defined by the parametric equations

find the time $t$ when the object changes direction in the $y$-direction.

Solve for the time(s) $t$ when the object crosses the $t$-axis (i.e., $y(t)=0$) for $y(t)=t^2+11t+4.$

Given the function $y(t) = 4 + t(11 + t)$, find the value of $t$ at which $y(t)$ attains its local minimum. Justify your answer.

Determine the interval(s) of $t$ for which the object moves upward (i.e. $\frac{dy}{dt} > 0$) given

Find the value(s) of $t$ for which the trajectory has a vertical tangent.

Find the time(s) $t$ when the trajectory $x(t)=3+t(9-t^2), \quad y(t)=4+t(11+t)$ has a horizontal tangent.