77. projectile motion a ball thrown follows a path described by y = x - 0.02x². (a) sketch a graph of the…

Explanation:

Step1: Identify the function type

The function $y = x - 0.02x^{2}$ is a quadratic - function of the form $y=ax^{2}+bx + c$ where $a=-0.02$, $b = 1$, and $c = 0$.

Step2: (a) Sketch the graph

The graph of a quadratic function $y = ax^{2}+bx + c$ is a parabola. Since $a=-0.02<0$, the parabola opens downwards. The $y$-intercept is at $(0,0)$ (when $x = 0$, $y=0$).

Step3: (b) Find the total horizontal distance

The ball hits the ground when $y = 0$. So we set $y=x - 0.02x^{2}=x(1 - 0.02x)=0$. We get $x = 0$ or $1-0.02x=0$. Solving $1-0.02x=0$ gives $x = 50$. The total horizontal distance is $50$ units.

Step4: (c) Find the $x$-value of the maximum

For a quadratic function $y = ax^{2}+bx + c$, the $x$-coordinate of the vertex (maximum for $a<0$) is given by $x=-\frac{b}{2a}$. Substituting $a=-0.02$ and $b = 1$ into $x=-\frac{b}{2a}$, we have $x=-\frac{1}{2\times(-0.02)} = 25$.

Step5: (d) Find the derivative

The instantaneous rate of change of $y$ with respect to $x$ is the derivative $y'$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, if $y=x - 0.02x^{2}$, then $y'=1-0.04x$. When $x = 0$, $y'(0)=1-0.04\times0 = 1$. When $x = 10$, $y'(10)=1-0.04\times10=1 - 0.4 = 0.6$. When $x = 25$, $y'(25)=1-0.04\times25=1 - 1 = 0$. When $x = 30$, $y'(30)=1-0.04\times30=1 - 1.2=-0.2$. When $x = 50$, $y'(50)=1-0.04\times50=1 - 2=-1$.

Step6: (e) Instantaneous rate at maximum

At the maximum height, the slope of the tangent line to the curve is $0$. From part (d), when $x = 25$ (the $x$-value of the maximum height), $y'=0$.

Answer:

(a) Sketch a downward - opening parabola passing through the origin. (b) $50$ (c) $25$ (d) $y'=1 - 0.04x$; $y'(0)=1$, $y'(10)=0.6$, $y'(25)=0$, $y'(30)=-0.2$, $y'(50)=-1$ (e) $0$