Polynomial Motion

1. **Problem Statement:** We have a polynomial function for the vertical displacement of a robotic arm nozzle: $$P(t) = t^4 - 8t^3 + 21t^2 - 18t + k$$ where $k$ is an unknown calibration constant. Given: - At $t=4$, $P(4) = 48$ mm. - The nozzle passes through the rest position ($P(t)=0$) at $t=2$ seconds. Tasks: (a) Comment on the degree of the motion path. (b) Find $k$. (c) Find all $t$ when $P(t)=0$. (d) Calculate $P(t)$ at $t=-5$, $t=0$, and $t=120$ seconds. (e) Describe behavior of $P(t)$ at $t-5$, $3+t$, and $2t+1$. 2. **Degree of the motion path:** The degree of a polynomial is the highest power of $t$. Here, the highest power is 4, so the degree is 4. This means the motion path is described by a quartic polynomial, which can have up to 4 real roots and can show complex motion patterns. 3. **Determine $k$ using $P(4) = 48$:** Substitute $t=4$: $$P(4) = 4^4 - 8(4)^3 + 21(4)^2 - 18(4) + k = 48$$ Calculate each term: $$4^4 = 256$$ $$8(4)^3 = 8 \times 64 = 512$$ $$21(4)^2 = 21 \times 16 = 336$$ $$18(4) = 72$$ So: $$256 - 512 + 336 - 72 + k = 48$$ Simplify: $$256 - 512 = -256$$ $$-256 + 336 = 80$$ $$80 - 72 = 8$$ So: $$8 + k = 48 \implies k = 48 - 8 = 40$$ 4. **Find all $t$ such that $P(t) = 0$ (nozzle passes rest position):** We know $P(2) = 0$ (given), so $t=2$ is a root. Substitute $k=40$ into $P(t)$: $$P(t) = t^4 - 8t^3 + 21t^2 - 18t + 40$$ Divide $P(t)$ by $(t-2)$ using polynomial division or synthetic division: Synthetic division setup: Coefficients: 1 (for $t^4$), -8, 21, -18, 40 Divide by root $t=2$: Bring down 1 Multiply 1*2=2, add to -8 = -6 Multiply -6*2 = -12, add to 21 = 9 Multiply 9*2=18, add to -18 = 0 Multiply 0*2=0, add to 40 = 40 (remainder) Since remainder is not zero, $t=2$ is not a root with $k=40$. This contradicts the problem statement. Re-examine step 3: The problem states the nozzle passes through 0 mm at $t=2$, so $P(2)=0$ must hold. Calculate $P(2)$ with $k=40$: $$2^4 - 8(2)^3 + 21(2)^2 - 18(2) + 40 = 16 - 64 + 84 - 36 + 40$$ Calculate stepwise: $$16 - 64 = -48$$ $$-48 + 84 = 36$$ $$36 - 36 = 0$$ $$0 + 40 = 40$$ So $P(2) = 40$, not 0. This means $k=40$ contradicts the condition $P(2)=0$. We must use the condition $P(2) = 0$ to find $k$ instead. Calculate $P(2)$: $$2^4 - 8(2)^3 + 21(2)^2 - 18(2) + k = 0$$ Calculate terms: $$16 - 64 + 84 - 36 + k = 0$$ Simplify: $$16 - 64 = -48$$ $$-48 + 84 = 36$$ $$36 - 36 = 0$$ So: $$0 + k = 0 \implies k = 0$$ Now use $k=0$ to check $P(4) = 48$: $$4^4 - 8(4)^3 + 21(4)^2 - 18(4) + 0 = ?$$ Calculate: $$256 - 512 + 336 - 72 = 8$$ But given $P(4) = 48$, so $k=0$ contradicts this. We have two conditions: - $P(4) = 48$ - $P(2) = 0$ Set up system: $$P(4) = 256 - 512 + 336 - 72 + k = 48$$ Simplify: $$8 + k = 48 \implies k = 40$$ $$P(2) = 16 - 64 + 84 - 36 + k = 0$$ Simplify: $$0 + k = 0 \implies k = 0$$ Contradiction means the problem's data is inconsistent or $k$ must be adjusted. Assuming $k$ is constant, the only way is to find $k$ satisfying both equations simultaneously. Set up equations: $$P(4) = 8 + k = 48 \implies k = 40$$ $$P(2) = 0 + k = 0 \implies k = 0$$ No single $k$ satisfies both. **Conclusion:** The problem likely means the nozzle passes through 0 mm at $t=2$ ignoring $k$, or $k$ is adjusted. We proceed assuming $k=40$ (from $P(4)=48$) and check roots. 5. **Find roots of $P(t) = 0$ with $k=40$:** $$t^4 - 8t^3 + 21t^2 - 18t + 40 = 0$$ Try rational roots using Rational Root Theorem: possible roots are factors of 40 over 1: $\pm1, \pm2, \pm4, \pm5, \pm8, \pm10, \pm20, \pm40$ Test $t=2$: $$16 - 64 + 84 - 36 + 40 = 40 \neq 0$$ Test $t=1$: $$1 - 8 + 21 - 18 + 40 = 36 \neq 0$$ Test $t=4$: $$256 - 512 + 336 - 72 + 40 = 48 \neq 0$$ Test $t=5$: $$625 - 1000 + 525 - 90 + 40 = 100 \neq 0$$ No easy rational roots. Use numerical methods or graphing to approximate roots. 6. **Calculate $P(t)$ at specified times:** Using $k=40$: - At $t=-5$: $$P(-5) = (-5)^4 - 8(-5)^3 + 21(-5)^2 - 18(-5) + 40$$ Calculate: $$625 + 1000 + 525 + 90 + 40 = 2280$$ - At $t=0$: $$P(0) = 0 - 0 + 0 - 0 + 40 = 40$$ - At $t=120$ (2 minutes = 120 seconds): Calculate dominant term: $$120^4 = 207360000$$ Other terms are smaller but significant: $$-8(120)^3 = -8(1728000) = -13824000$$ $$21(120)^2 = 21(14400) = 302400$$ $$-18(120) = -2160$$ Sum: $$207360000 - 13824000 + 302400 - 2160 + 40 = 194,444,280$$ 7. **Behavior at checkpoints:** - At $t-5$: substitute $t-5$ into $P(t)$: $$P(t-5) = (t-5)^4 - 8(t-5)^3 + 21(t-5)^2 - 18(t-5) + 40$$ This shifts the graph 5 units to the right. - At $3+t$: substitute $3+t$: $$P(3+t) = (3+t)^4 - 8(3+t)^3 + 21(3+t)^2 - 18(3+t) + 40$$ This shifts the graph 3 units to the left. - At $2t+1$: substitute $2t+1$: $$P(2t+1) = (2t+1)^4 - 8(2t+1)^3 + 21(2t+1)^2 - 18(2t+1) + 40$$ This compresses the graph horizontally by factor 2 and shifts it left by 1. **Final answers:** - Degree: 4 - $k = 40$ - Roots require numerical methods (no simple rational roots) - Heights: - $P(-5) = 2280$ - $P(0) = 40$ - $P(120) = 194,444,280$ - Behavior at checkpoints: horizontal shifts and scaling as above.