Question

Find dw/dt at t=(pi)/2
where w=x2+y2+2x+3y , x=cos t , y=sin t

Answer 1

Answer on Question #41381– Math - Integral Calculus

Question:

Find dw/dt at $t = (pi)/2$

where $w = x^2 + y^2 + 2x + 3y$, $x = \cos t$, $y = \sin t$

Solution:

$$\frac{dw}{dt} = \frac{d}{dt} (x^2 + y^2 + 2x + 3y) = 2x * x' + 2y * y' + 2x' + 3y'.$$

Since, $x' = -\sin t$, and $y' = \cos t$, we get

$$\frac{dw}{dt} = 2 \cos t * (-\sin t) + 2 \sin t * \cos t + 2(-\sin t) + 3 \cos t = \\ = -2 \cos t * \sin t + 2 \sin t * \cos t - 2 \sin t + 3 \cos t = -2 \sin t + 3 \cos t.$$

Thus, $\frac{dw}{dt} \left( \frac{\pi}{2} \right) = -2 \sin \frac{\pi}{2} + 3 \cos \frac{\pi}{2} = -2 + 0 = -2$.

Answer:

$$\frac{dw}{dt} \left( \frac{\pi}{2} \right) = -2.$$

http://www.AssignmentExpert.com