1) If y = 2x3 - 4x and dx, dt equals 4 , find dy/ dt when x = 1.
2) The area A = πr^2 of a circular water ripple changes with the radius. At what rate does the area change with respect to the radius when r = 4ft?
3) A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x^2h cm^3. Find the rate at which the volume of the box is changing when the edge length of the base is 10 cm, the edge length of the base is increasing at a rate of 3 cm/min, the height of the box is 5 cm, and the height is decreasing at a rate of 1 cm/min.
4) The height of a cylinder with a fixed radius of 10 cm is increasing at the rate of 0.5 cm/min. Find the rate of change of the volume of the cylinder (with respect to time) when the height is 30cm.
Where are your attempts to do these? Where did you get stuck? I already took the subject, do not need practice. They are all really the same. For example the last one:
r = 10
V = pi r^2 h = 100pi h
100 pi is a constant so
dV/dt = 100 pi * dh/dt
or the second one (the first one has a typo)
dA/dt = dA/dr * dr/dt = 2 pi r * dr/dt
in other words the area increase is the circumference times dr
Deondre = James
=(
Never would have guessed :)
1) To find dy/dt when x = 1, we need to differentiate the equation y = 2x^3 - 4x with respect to time (t), using the chain rule. The chain rule states that if y = f(g(t)), then the derivative of y with respect to t is given by dy/dt = f'(g(t)) * g'(t).
In this case, f(u) = u^3 - 4u and g(t) = 2t. We differentiate f(u) with respect to u to get f'(u) = 3u^2 - 4. Also, g'(t) = 2.
Now, we plug in x = 1 and dx/dt = 4 into the chain rule formula to find dy/dt.
dy/dt = f'(g(t)) * g'(t)
= f'(g(1)) * g'(4)
= f'(2) * 2
= (3(2)^2 - 4) * 2
= (3(4) - 4) * 2
= (12 - 4) * 2
= 8 * 2
= 16
Therefore, dy/dt = 16 when x = 1.
2) The area A = πr^2 of a circular water ripple with respect to the radius r can be found by differentiating the equation A = πr^2 with respect to r.
We differentiate A = πr^2 with respect to r to get:
dA/dr = 2πr
Now, we plug in r = 4ft into the equation to find the rate at which the area changes with respect to the radius.
dA/dr = 2πr
= 2π(4)
= 8π
Therefore, when r = 4ft, the area changes at a rate of 8π square feet per unit change in radius.
3) The volume V = x^2h of the rectangular box with respect to the edge length x can be found by differentiating the equation V = x^2h with respect to x.
We differentiate V = x^2h with respect to x to get:
dV/dx = 2xh
Now, we plug in x = 10cm, dx/dt = 3cm/min, h = 5cm, and dh/dt = -1cm/min into the equation to find the rate at which the volume changes.
dV/dt = dV/dx * dx/dt
= (2xh) * (dx/dt)
= (2(10)(5)) * 3
= 100 * 3
= 300
Therefore, the rate at which the volume of the box is changing is 300 cm^3/min.
4) The volume V of a cylinder with respect to the height h can be found by differentiating the formula V = πr^2h, where r is the fixed radius.
We differentiate V = πr^2h with respect to h to get:
dV/dh = πr^2
Now, we plug in r = 10cm and dh/dt = 0.5cm/min into the equation to find the rate of change of the volume.
dV/dt = dV/dh * dh/dt
= (πr^2) * (dh/dt)
= (π(10)^2) * 0.5
= (π(100)) * 0.5
= (100π) * 0.5
= 50π
Therefore, when the height is 30cm, the rate of change of the volume of the cylinder is 50π cm^3/min.