question 3 (continued)\n3.3 determine the gradient of the curve $y = \\frac{2x}{x^{2}-5}$ at the point $(2…

question 3 (continued)\n3.3 determine the gradient of the curve $y = \\frac{2x}{x^{2}-5}$ at the point $(2, - 4)$\n3.4 a particle has a displacement $s$, given by $s = 30t + 27t^{2}-3t^{3}$ meters, where $t$ is in seconds. determine the time at which the acceleration will be zero.

question 3 (continued)\n3.3 determine the gradient of the curve $y = \\frac{2x}{x^{2}-5}$ at the point $(2, - 4)$\n3.4 a particle has a displacement $s$, given by $s = 30t + 27t^{2}-3t^{3}$ meters, where $t$ is in seconds. determine the time at which the acceleration will be zero.

Answer

Explanation:

Step1: Find the derivative of $y = \frac{2x}{x^{2}-5}$ using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 2x$, $u^\prime=2$, $v=x^{2}-5$, $v^\prime = 2x$. So, $y^\prime=\frac{2(x^{2}-5)-2x(2x)}{(x^{2}-5)^{2}}=\frac{2x^{2}-10 - 4x^{2}}{(x^{2}-5)^{2}}=\frac{-2x^{2}-10}{(x^{2}-5)^{2}}$.

Step2: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $y^\prime$: $y^\prime|_{x = 2}=\frac{-2(2)^{2}-10}{(2^{2}-5)^{2}}=\frac{-8 - 10}{(4 - 5)^{2}}=\frac{-18}{1}=-18$.

Step3: Find the velocity and acceleration functions for the particle

The displacement function is $s=30t + 27t^{2}-3t^{3}$. The velocity function $v=\frac{ds}{dt}=30 + 54t-9t^{2}$. The acceleration function $a=\frac{dv}{dt}=54 - 18t$.

Step4: Set the acceleration equal to zero and solve for $t$

Set $a = 0$, so $54-18t=0$. Add $18t$ to both sides: $54 = 18t$. Then divide both sides by 18 to get $t = 3$ seconds.

Answer:

3.3: - 18 3.4: 3 seconds