# Thread: How is this done? 2 questions - Curves + Derivatives and Related Rates

1. ## How is this done? 2 questions - Curves + Derivatives and Related Rates

I'm doing some Calculus homework, but these two problems are a bit confusing. Thanks for taking the time to look:

Consider the curve defined by 2y^3 + 6x^(2)y - 12x^2 + 6y = 1

1. Find dy/dx (Found using implicit differentiation; (4x-2xy)/(x^2+y^2+1))
2. Write an equation of each horizontal tangent line to the curve (Don't understand)
3. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates at point P. (Also don't understand)
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Here is the second question:
Ship A is traveling due west toward Blunt Rock at a speed of 15 km/hr. Ship B is traveling due north away from Blunt Rock at a speed of 10 km/hr. Let x be the distance between Ship A and Blunt Rock at time t, and let y be the distance between Ship B and Blunt Rock at time t.

1. Find the distance, in km, between Ship A and Ship B when x=4 km and y=3 km (seems simple, use pythagorean theorem to get hypotenuse = 5 km)
2. Find the rate of change, in km/hr, of the distance between the two ships when x=4km and y=3km. (Think I got this, using implicit on pytahgorean theorem I get -6km/hr)
3. Let theta be the angle [at ship A, between the x axis and hypotenuse]. Find the rate of change of theta, in radians per hour, when x=4km and y=3km. (Not sure how this is done)

Thank you in advance

2. Originally Posted by Harryhit4
Sorry if my title sucked! I'm doing some Calculus homework, but these two problems are a bit confusing. Thanks for taking the time to look:

Consider the curve defined by 2y^3 + 6x^(2)y - 12x^2 + 6y = 1

1. Find dy/dx (Found using implicit differentiation); (4x-2xy)/(x^2+y^2+1))
2. Write an equation of each horizontal tangent line to the curve (Don't understand)
3. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates at point P. (Also don't understand)
--

Thank you in advance
2. The equation of a horizontal line is of the form y=k.

If it is a tangent to the curve, the derivative must be zero,
since the derivative formulates the slope of the tangent,
hence for what values of x or y is the numerator of the derivative zero?

3. This is the line y=-x.

The slope of a line is it's derivative.

Hence the derivative is -1, and you can substitute y=-x into the derivative to solve for x, and thus y.

3. Originally Posted by Harryhit4
Here is the second question:
Ship A is traveling due west toward Blunt Rock at a speed of 15 km/hr. Ship B is traveling due north away from Blunt Rock at a speed of 10 km/hr. Let x be the distance between Ship A and Blunt Rock at time t, and let y be the distance between Ship B and Blunt Rock at time t.

1. Find the distance, in km, between Ship A and Ship B when x=4 km and y=3 km (seems simple, use pythagorean theorem to get hypotenuse = 5 km)
2. Find the rate of change, in km/hr, of the distance between the two ships when x=4km and y=3km. (Think I got this, using implicit on pytahgorean theorem I get -6km/hr)
3. Let theta be the angle [at ship A, between the x axis and hypotenuse]. Find the rate of change of theta, in radians per hour, when x=4km and y=3km. (Not sure how this is done)

Thank you in advance
2. On your drawing, ship A is moving to left, towards the 90 degree corner at 15km/hr.

Ship B is moving up, away from the corner at 10km/hr.

$\displaystyle\frac{dx}{dt}=-15km/hr$

$\displaystyle\frac{dy}{dt}=10km/hr$

$\frac{dx}{dt}$ is negative as it is moving towards the triangle's right-angle,
instead of moving away from it.

hence, we ought to expect a negative value for the rate of change of the hypotenuse,
since ship A is going a faster rate of knots than ship B.

$L^2=x^2+y^2$

$\displaystyle\frac{d}{dt}L^2=\frac{d}{dt}x^2+\frac {d}{dt}y^2$

$2L\frac{dL}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$

You have the values for L, x, y, dx/dt and dy/dt, hence dL/dt can be solved for.

4. Hello, Harryhit4!

$\text{Ship A is traveling due west toward Blunt Rock at a speed of 15 kph.}$
$\text{Ship B is traveling due north away from Blunt Rock at a speed of 10 kph.}$
$\text{Let }x\text{ be the distance between Ship A and Blunt Rock at time }t.$
$\text{Let }y\text{ be the distance between Ship B and Blunt Rock at time }t.$

$\text{1. Find the distance, in km, between Ship A and Ship B}$
. . . $\text{when }x=4\text{ and }y=3.$
$\text{(seems simple, use Pythagorus to get hypotenuse = 5 km)}$ .Good!

$\text{2. Find the rate of change, in kph, of the distance between the ships}$
. . . $\text{when }x=4 \text{ and }y=3.$
$\text{(Think I got this, using implicit on Pythagorus I get: -6 kph)}$ .Yes!

$\text{3. Let }\theta\text{ be the angle at ship A, between the }x\text{-axis and hypotenuse.}$
$\text{Find the rate of change of }\theta\text{, in radians/hr, when }x =4\text{ and }y=3.$
Code:
    B *
: |  *
: |     *
y |        *
: |           *
: |            @ *
R *-----------------*
: - - -  x  - - - A

Blunt Rock is at $\,R.$

$\text{Ship A is at }\,A\!:\;x = AR,\;\dfrac{dx}{dt} = \text{-}6\text{ kph}$

$\text{Ship B is at }\,B\!:\;y = BR,\;\dfrac{dy}{dt} = 10\text{ kph}$

$\,\theta \,=\,\angle A$

Note: .when $x = 4,\;y = 3\!:\;\cos\theta \:=\:\frac{4}{5}$

$\text{We have: }\;\tan\theta \:=\:\dfrac{y}{x}$

$\text{Then: }\;\sec^2\!\theta\,\dfrac{d\theta}{dt} \;=\;\dfrac{x\frac{dy}{dt} - y \frac{dx}{dt}}{x^2}$

. . . . . . . . . . $\dfrac{d\theta}{dt} \;=\;\dfrac{x\frac{dy}{dt} - y\frac{x}{dt}}{x^2}\cos^2\!\theta$

. . . . . . . . . . $\dfrac{d\theta}{dt} \;=\;\dfrac{4(10) - 3(\text{-}15)}{4^2}\cdot \left(\frac{4}{5}\right)^2$

. . . . . . . . . . $\displaystyle \frac{d\theta}{dt} \;=\;\frac{85}{16}\cdot\frac{16}{25} \;=\;\frac{17}{5} \;=\;3\tfrac{2}{5}\text{ radians/hour}$