Find the equation of the line tangent to the circle (x-3)^2+y^2=7 at the point(4

Question

Find the equation of the line tangent to the circle (x-3)^2+y^2=7 at the point(4, squareroot 6). A glass in the shape of an inverted cone is being filled with liquid at a rate of 120 xm^2/mm Given that it has a base radius of 10cm and a height of 8 cm, how is the liquid level changing when the height is 5cm? Two dogs are on opposite ends of a street that runs east-west, 100 meters apart, Coco heads north at 10 meters per minute and Buffy also heads north at 12 meters per minute. How is the distance between them changing after half an hour? A camera is positioned 160 ft across the street from your mailbox, perpendicular to the sidewalk you are on. It rotates to follow you as you walk toward your mailbox at a speed of 3 feet per second. Describe the change in the angle of rotation of the cameral when you are 20 feet from the mailbox.

Explanation / Answer

3) Take derivative of given equation on both sides with respect to x, we get:

2(x-3) + 2ydy/dx = 0

=> (x-3) = -ydy/dx

Now for point x=4, we have y=sqrt6

So (4-3) = -sqrt(6) dy/dx

=> -1/sqrt(6) = dy/dx = slope of tangent line to circle at given point

So we have equation of line given by:

y-y1 = m (x-x1)

=> y-sqrt(6) = -1/sqrt(6) (x-4)

=> y-sqrt6 = -x/sqrt6+4/sqrt6

=> y = -x/sqrt6+4/sqrt6+sqrt6 = -x/sqrt(6)+10/sqrt(6) = (-x+10)/sqrt(6)

Hence y = (-x+10)/sqrt(6) is required equation of tangent line

4) Let V be the volume of the cone, so we are given dV/dt = 120 cm3/min

Now we need to know dh/dt

By using similar traingles, we get:

8/10 = h/r => h/r = 4/5 => 5h = 4r=> r = (5/4) h

Also, we have: V = (1/3)pi(5/4h)2h = (25/48)pih3

Taking derivative both sides, we get:

dV/dt = (25/48)pi(3h2)dh/dt

Now when h=5,

So we get:
120 = (25/16)(25)2dh/dt

=> dh/dt = 120(16)/(25)3 = 0.12288 cm/min

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