Implicit Differentiation
Section 2.5 (B) Question 1(a)
Find dy/dx of the given functions by implicit differentiation.
x2 y3 − y = 4x2
Differentiate w.r.t x
2xy3 + 3x2y2y' − y' = 8x
3x2y2y' − y' = 8x − 2xy3
(3x2y2 − 1)y' = 8x − 2xy3
y' = (8x − 2xy3)/(3x2y2 − 1)
Section 2.6 (B) Question 2
A baseball diamond is a square with each side being 90 feet. A player runs toward first base from home plate at a speed of 30 feet per second. How fast is his distance from second base decreasing when he is 2/3 of the way to first base?
Known:
y = 90
dx/dt = −30
x = 1/3 · 90 = 30
dz/dt =?
Find Missing Side:
x2 + y2 = z2
302 + 902= z2
z2 = 8100 + 900
z = √9000
Solve:
d/dt(x2 + 902 ) = d/dt(z2)
2x · dx/dt = 2z · dz/dt
x · dx/dt = z · dz/dt
30(−30) = √ 9000 · dz/dt
dz/dt = −900 √ 9000 ≈ −9.487
Section 2.6 (B) Question 1
A cylindrical water tank with a radius of 2 ft is being filled at a rate of 4 ft3/min. How fast is the level of the water in the tank rising when the tank is half full? Use the fact that the volume of a cylinder is given by the equation V = pir2h.
dV/dt = 4, r = 2
V = pir2h
V = 4pih
dV/dt = 4pi dh/dt
4 = 4pi dh/dt
dh/dt = 1/pi ft/min
Section 2.5 (B) Question 4
Find the equation of the tangent line to the equation
xcos(y) = 1 at the point (2, pi/3)
Equation of the tangent line at the point (2, pi/3) is given by:
y − (pi/3) = dy/dx (x −2)
Use use implicit differentiation to find the slope to the given equation: xcos(y) = 1.
Differentiate w.r.t x
cos(y) − xsin(y)y' = 0
−xsin(y)y' = − cos(y)
y'= cos(y)/ xsin(y)
At the point, (2, pi/3), Slope is given by: y' = 1/(2√3). Thus, the equation of the line is:
y − (pi/3) = 1/(2√ 3)(x-2)
Section 2.6 (B) Question 4
A firework is shot into the air, and a spectator is watching from 25 meters away. When the firework is 75 meters above ground, its speed is 150 meters/second. At that point in time, what is the rate of change of the angle of elevation, i.e., the angle between the ground and the spectator’s line of sight to the firework?
Know:
y = 75
dy/dt = 150
dθ/dt =?
Solve:
tan θ = y/25
tan θ = 75/25
θ ≈ 1.249
d/dt(tan θ) = d/dt (y/25)
sec2θ · dθ/dt = (dy/dt)/25
sec2 1.249 · dθ/dt = 150/25
dθ/dt = 6 · cos2 * 1.249
dθ/dt ≈ .6 radians/sec