A particle travels along the x-axis such that its position is given by x(t)=t^(1.9)sin(t^(2)). What is the velocity of the particle at time t=0.5 ? You may use a calculator and round your answer to the nearest thousandth. Answer:

Differentiate Position Function: To find the velocity of the particle, we need to differentiate the position function x(t) with respect to time t. The velocity function v(t) is the derivative of the position function x(t).Apply Product Rule: The position function is given by x(t) = t^(1.9)sin(t^(2)). We will use the product rule for differentiation, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.Evaluate at t=0.5: Differentiate t^(1.9) with respect to t to get 1.9t^(0.9).Calculate Approximations: Differentiate sin(t^(2)) with respect to t using the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. The derivative of sin(u) with respect to u is cos(u), and the derivative of t^(2) with respect to t is 2t. Therefore, the derivative of sin(t^(2)) with respect to t is 2t*cos(t^(2)).Perform Calculations: Now, apply the product rule: the derivative of x(t) is (1.9t^(0.9))sin(t^(2)) + t^(1.9)(2t*cos(t^(2))).Round to Nearest Thousandth: Simplify the expression to get the velocity function v(t): v(t) = 1.9t^(0.9)sin(t^(2)) + 2t^(2.9)cos(t^(2)).Evaluate the velocity function at t=0.5. We plug in t=0.5 into the velocity function v(t) to get v(0.5) = 1.9(0.5)^(0.9)sin((0.5)^(2)) + 2(0.5)^(2.9)cos((0.5)^(2)).Using a calculator, we find that (0.5)^(0.9) ≈ 0.574, sin((0.5)^(2)) = sin(0.25) ≈ 0.247, (0.5)^(2.9) ≈ 0.176, and cos((0.5)^(2)) = cos(0.25) ≈ 0.968.Now, calculate v(0.5) using the approximations: v(0.5) ≈ 1.9 * 0.574 * 0.247 + 2 * 0.176 * 0.968.Perform the calculations: v(0.5) ≈ 1.9 * 0.574 * 0.247 + 2 * 0.176 * 0.968 ≈ 0.267 + 0.341 ≈ 0.608.Round the answer to the nearest thousandth: v(0.5) ≈ 0.608.

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A particle travels along the
x-axis such that its position is given by
x(t)=t^(1.9)sin(t^(2)). What is the velocity of the particle at time
t=0.5 ? You may use a calculator and round your answer to the nearest thousandth.
Answer:

A particle travels along the $x$ -axis such that its position is given by $x(t)=t^{1.9} \sin \left(t^{2}\right)$ . What is the velocity of the particle at time $t=0.5$ ? You may use a calculator and round your answer to the nearest thousandth. $\newline$ Answer:

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Calculus

Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A particle travels along the

x

-axis such that its position is given by

x(t)=t^{1.9} \sin \left(t^{2}\right)

. What is the velocity of the particle at time

t=0.5

? You may use a calculator and round your answer to the nearest thousandth.

\newline

Answer:

Differentiate Position Function: To find the velocity of the particle, we need to differentiate the position function $x(t)$ with respect to time $t$ . The velocity function $v(t)$ is the derivative of the position function $x(t)$ .
Apply Product Rule: The position function is given by $x(t) = t^{1.9}\sin(t^{2})$ . We will use the product rule for differentiation, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
Evaluate at $t=0.5$ : Differentiate $t^{1.9}$ with respect to $t$ to get $1.9t^{0.9}$ .
Calculate Approximations: Differentiate $\sin(t^{2})$ with respect to $t$ using the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ , and the derivative of $t^{2}$ with respect to $t$ is $2t$ . Therefore, the derivative of $\sin(t^{2})$ with respect to $t$ is $t$ $0$ .
Perform Calculations: Now, apply the product rule: the derivative of $x(t)$ is $(1.9t^{0.9})\sin(t^{2}) + t^{1.9}(2t\cos(t^{2}))$ .
Round to Nearest Thousandth: Simplify the expression to get the velocity function $v(t)$ : $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ .
Round to Nearest Thousandth: Simplify the expression to get the velocity function $v(t)$ : $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ .Evaluate the velocity function at $t=0.5$ . We plug in $t=0.5$ into the velocity function $v(t)$ to get $v(0.5) = 1.9(0.5)^{0.9}\sin((0.5)^{2}) + 2(0.5)^{2.9}\cos((0.5)^{2})$ .
Round to Nearest Thousandth: Simplify the expression to get the velocity function $v(t)$ : $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ .Evaluate the velocity function at $t=0.5$ . We plug in $t=0.5$ into the velocity function $v(t)$ to get $v(0.5) = 1.9(0.5)^{0.9}\sin((0.5)^{2}) + 2(0.5)^{2.9}\cos((0.5)^{2})$ .Using a calculator, we find that $(0.5)^{0.9} \approx 0.574$ , $\sin((0.5)^{2}) = \sin(0.25) \approx 0.247$ , $(0.5)^{2.9} \approx 0.176$ , and $\cos((0.5)^{2}) = \cos(0.25) \approx 0.968$ .
Round to Nearest Thousandth: Simplify the expression to get the velocity function $v(t)$ : $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ .Evaluate the velocity function at $t=0.5$ . We plug in $t=0.5$ into the velocity function $v(t)$ to get $v(0.5) = 1.9(0.5)^{0.9}\sin((0.5)^{2}) + 2(0.5)^{2.9}\cos((0.5)^{2})$ .Using a calculator, we find that $(0.5)^{0.9} \approx 0.574$ , $\sin((0.5)^{2}) = \sin(0.25) \approx 0.247$ , $(0.5)^{2.9} \approx 0.176$ , and $\cos((0.5)^{2}) = \cos(0.25) \approx 0.968$ .Now, calculate $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $0$ using the approximations: $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $1$ .
Round to Nearest Thousandth: Simplify the expression to get the velocity function $v(t)$ : $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ .Evaluate the velocity function at $t=0.5$ . We plug in $t=0.5$ into the velocity function $v(t)$ to get $v(0.5) = 1.9(0.5)^{0.9}\sin((0.5)^{2}) + 2(0.5)^{2.9}\cos((0.5)^{2})$ .Using a calculator, we find that $(0.5)^{0.9} \approx 0.574$ , $\sin((0.5)^{2}) = \sin(0.25) \approx 0.247$ , $(0.5)^{2.9} \approx 0.176$ , and $\cos((0.5)^{2}) = \cos(0.25) \approx 0.968$ .Now, calculate $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $0$ using the approximations: $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $1$ .Perform the calculations: $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $2$ .
Round to Nearest Thousandth: Simplify the expression to get the velocity function $v(t)$ : $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ .Evaluate the velocity function at $t=0.5$ . We plug in $t=0.5$ into the velocity function $v(t)$ to get $v(0.5) = 1.9(0.5)^{0.9}\sin((0.5)^{2}) + 2(0.5)^{2.9}\cos((0.5)^{2})$ .Using a calculator, we find that $(0.5)^{0.9} \approx 0.574$ , $\sin((0.5)^{2}) = \sin(0.25) \approx 0.247$ , $(0.5)^{2.9} \approx 0.176$ , and $\cos((0.5)^{2}) = \cos(0.25) \approx 0.968$ .Now, calculate $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $0$ using the approximations: $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $1$ .Perform the calculations: $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $2$ .Round the answer to the nearest thousandth: $v(t) = 1.9t^{0.9}\sin(t^{2}) + 2t^{2.9}\cos(t^{2})$ $3$ .