Find the derivative r '(t) of the vector function r(t).
<t cos 5t , t2, t sin 5t >
To find the derivative of the vector function r(t) = <t cos 5t , t^2, t sin 5t>, we need to take the derivative of each component of the vector separately.
The derivative of the first component, t cos 5t, can be found using the product rule. The product rule states that if we have a function f(t) multiplied by g(t), the derivative of the product is given by:
(f(t)g(t))' = f'(t)g(t) + f(t)g'(t)
Applying the product rule to our first component, we have:
f(t) = t
g(t) = cos 5t
f'(t) = 1 (since the derivative of t is 1)
g'(t) = -5 sin 5t (since the derivative of cos 5t is -5 sin 5t)
Using the product rule formula, the derivative of the first component is then:
f'(t)g(t) + f(t)g'(t) = 1 * cos 5t + t * (-5 sin 5t) = cos 5t - 5t sin 5t
The second component of the vector is t^2. To find its derivative, we can use the power rule, which states that the derivative of t^n is given by:
d(t^n) / dt = n * t^(n-1)
For the second component, n = 2, so applying the power rule gives us:
d(t^2) / dt = 2 * t^(2-1) = 2t
Finally, the third component of the vector is t sin 5t. Again, we can use the product rule to find its derivative.
f(t) = t
g(t) = sin 5t
f'(t) = 1
g'(t) = 5 cos 5t
Using the product rule formula, the derivative of the third component is:
f'(t)g(t) + f(t)g'(t) = 1 * sin 5t + t * (5 cos 5t) = sin 5t + 5t cos 5t
Putting it all together, the derivative of the vector function r(t) = <t cos 5t , t^2, t sin 5t> is:
r'(t) = <cos 5t - 5t sin 5t, 2t, sin 5t + 5t cos 5t>