the vector function r(t)=< 5 sin t, 3 sec t > , represents the position of a particle at time t. find the velocity v (t) and acceleration a (t).
I don't know how to solve it without the t given. Please help
v(t) = dr/dt = <5cost, 3sect tant>
a(t) = dv/dt = <-5sint, 3sect(2sec^2t-1)>
just do normal derivatives of trig functions
Find the area of the shape shown below.
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To find the velocity vector, you need to find the derivative of the position vector. Given the position vector r(t) = <5 sin t, 3 sec t>, differentiate each component with respect to t:
r'(t) = <d/dt (5 sin t), d/dt (3 sec t)>
To find the derivative of 5 sin t, you can use the chain rule. The derivative of sin t is cos t. Since we have 5 multiplied by sin t, the constant 5 remains unaffected:
d/dt (5 sin t) = 5 * (cos t)
To find the derivative of 3 sec t, we need to use the derivative of the secant function. The derivative of sec t is sec t * tan t. Again, the constant 3 remains the same:
d/dt (3 sec t) = 3 * (sec t * tan t)
Putting it all together, the velocity vector v(t) is:
v(t) = <5cos t, 3sec t * tan t>
Now, to find the acceleration vector, you need to take the derivative of the velocity vector:
a(t) = d/dt (v(t))
Differentiate each component of v(t) with respect to t:
a(t) = <d/dt (5cos t), d/dt (3sec t * tan t)>
The derivative of 5cos t is -5sin t:
d/dt (5cos t) = -5sin t
The derivative of 3sec t * tan t involves the product rule. The product rule states that if you have two functions u(t) and v(t), then the derivative of their product is:
d/dt (u(t) * v(t)) = u'(t) * v(t) + u(t) * v'(t)
Using this rule, we have:
d/dt (3sec t * tan t) = 3 * (tan t * sec t) + 3sec t * (sec^2 t)
Simplifying, we get:
d/dt (3sec t * tan t) = 3sec t * (tan t + sec^2 t)
Therefore, the acceleration vector a(t) is:
a(t) = <-5sin t, 3sec t * (tan t + sec^2 t)>
This is the velocity and acceleration of the particle at any given time t, without a specific value of t being required.