In order to find the velocity of any point given a position function, we must first find the derivative of the position function, \(\displaystyle p'(t)= v(t)\).

This is the position function: \(\displaystyle p(t) = 3t - \cos t + \tan t \sin t\)

The derivative gives us the velocity function: \(\displaystyle v(t) = \cos t \tan t + \sec^ 2t \sin t + \sin t +3\) 

You can then find the velocity of a given point by substituting it in for the variable: \(\displaystyle v(1) = \cos (1) \tan (1) + \sec^ 2(1) \sin (1) + \sin (1) +3\)

Therefore, the velocity at \(\displaystyle t= 1\) is: \(\displaystyle 7.6\)

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, \(\displaystyle p'(t)= v(t)\).

This is the position function: \(\displaystyle p(t) = ln(t) + 2e^t\)

The derivative gives us the velocity function: \(\displaystyle v(t) = \frac{1}{t} + 2e^t\) 

You can then find the velocity of a given point by substituting it in for the variable: \(\displaystyle v(0) = \frac{1}{(0)} + 2e^{(0)}\)

Therefore, the velocity at \(\displaystyle t= 0\) is: \(\displaystyle undefined\)

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, \(\displaystyle p'(t)= v(t)\).

This is the position function: \(\displaystyle p(t) = e^t + 3t^{(\frac{1}{3})} +4t^6\)

The derivative gives us the velocity function: \(\displaystyle v(t) = e^t + t^{-\frac{2}{3}} + 24t^5\)

You can then find the velocity of a given point by substituting it in for the variable: \(\displaystyle v(2) = e^{(2)} + (2)^{-\frac{2}{3}} + 24(2)^5\)

Therefore, the velocity at \(\displaystyle t=2\) is: \(\displaystyle 776.02\)

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, \(\displaystyle p'(t)= v(t)\).

This is the position function: \(\displaystyle p(t) = (4t^2+2t)^3 - 5\)

The derivative gives us the velocity function: \(\displaystyle v(t) = 3(4t^2+2t)^2 (8t+2)\) 

You can then find the velocity of a given point by substituting it in for the variable: \(\displaystyle v(3) = 3(4(3)^2+2(3))^2 (8(3)+2)\)

Therefore, the velocity at \(\displaystyle t= 3\) is: \(\displaystyle 137,592\)

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: \(\displaystyle p'(t)= v(t)\)

In this case, the position function is: \(\displaystyle p(t)= 2t^{\frac{1}{2}} + 3e^t +4\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= t^{-\frac{1}{2}} + 3e^t\)

Finally, to find the velocity, substitute \(\displaystyle t=4\) into the velocity function: \(\displaystyle v(4)= (4)^{-\frac{1}{2}} + 3e^{(4)}\)

Therefore, the answer is: \(\displaystyle 165.8\)

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: \(\displaystyle p'(t)= v(t)\)

In this case, the position function is: \(\displaystyle p(t)= \frac{1}{3}t^3 + ln(t) - 23t\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= t^2 + \frac{1}{t} - 23\)

Finally, to find the velocity, substitute \(\displaystyle t=3\) into the velocity function: \(\displaystyle v(3)= (3)^2 + \frac{1}{(3)} - 23\)

Therefore, the answer is: \(\displaystyle -13.67\)

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: \(\displaystyle p'(t)= v(t)\)

In this case, the position function is: \(\displaystyle p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= -t^{-\frac{7}{6}} + 20t^4 - 8t\)

Finally, to find the velocity, substitute \(\displaystyle t=1\) into the velocity function: \(\displaystyle v(1)= -(1)^{-\frac{7}{6}} + 20(1)^4 - 8(1)\)

Therefore, the answer is: \(\displaystyle 11\)

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: \(\displaystyle p'(t)= v(t)\)

In this case, the position function is: \(\displaystyle p(t)= -\cos t + 3t^2 + 6ln(t)\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= \sin t + 6t + \frac{6}{t}\)

Finally, to find the velocity, substitute \(\displaystyle t=\pi\) into the velocity function: \(\displaystyle v(\pi)= \sin (\pi) + 6(\pi) + \frac{6}{(\pi)}\)

Therefore, the answer is: \(\displaystyle 20.8\)

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: \(\displaystyle p'(t)= v(t)\)

In this case, the position function is: \(\displaystyle p(t)= -\cos t + \sin t - 45e^t + 5t^2\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= \sin t + \cos t - 45e^t + 10t\)

Finally, to find the velocity, substitute \(\displaystyle t=0\) into the velocity function: \(\displaystyle v(0)= \sin (0) + \cos(0) - 45e^{(0)} + 10(0)\)

Therefore, the answer is: \(\displaystyle -44\)

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

\(\displaystyle y_n=y_{n+1} +\Delta x f'(x_n,y_n)\)

In the case of this problem, this can be rewritten as:

\(\displaystyle p(t_n)=p(t_{n+1}) +\Delta t v(t_n)\)

To calculate the step size find the difference between the final and initial value of \(\displaystyle t\) and divide by the number of steps to be used:

\(\displaystyle \Delta t = \frac{t_f-t_i}{Steps}\)

For this problem, we are told \(\displaystyle v(t)=\frac{tan(t)}{t^3}\) \(\displaystyle p(1)=0\)

Knowing this, we may take the steps to estimate our function value at our desired \(\displaystyle t\) value:

\(\displaystyle \Delta t = \frac{1.6-1}{3}=0.2\)

\(\displaystyle p_0=0;t_0=1\)

\(\displaystyle p_1=0+(0.2)(\frac{tan(1)}{1^3})=0.311\)

\(\displaystyle p_2=0.311+(0.2)(\frac{tan(1.2)}{1.2^3})=0.609\)

\(\displaystyle p_3=0.609+(0.2)(\frac{tan(1.4)}{1.4^3})=1.032\)