# AP Calculus help

AP Calculus help!!!

## Homework Statement

consider y^2=2+xy

a. show that dy/dx = y/2y-x
b. find all pints (x,y) on the curve where the line tangent to the curve has slope 1/2
c. show that there are no points (x,y) on the curve where the line tangent to the curve is horizontal
d. Let x and y be functions of time t that are related by the equation y^2=2+xy. At time t=5, the value of y is 3 and dy/dt=6. Find the value of dx/dt at time t=5.

## Homework Equations

chain rule, implicit differentiation

## The Attempt at a Solution

for this did part a, but I am really confused for b, c and d

Last edited:

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Parts B and C:
some hints:
How would you interpret the derivative of a function at a point geometrically?
Notice also that in part a you have shown dy/dx to be a function of x and y. How can you relate this function to the geometrical interpretation of the derivative?

dynamicsolo
Homework Helper

## Homework Statement

consider y^2=2+xy

a. show that dy/dx = y/2y-x
b. find all pints (x,y) on the curve where the line tangent to the curve has slope 1/2
Set dy/dx = 1/2 . What do you find?

c. show that there are no points (x,y) on the curve where the line tangent to the curve is horizontal
Set dy/dx = 0. What do you find? (Do you find anything?)

d. Let x and y be functions of time t that are related by the equation y^2=2+xy. At time t=5, the value of y is 3 and dy/dt=6. Find the value of dx/dt at time t=5.
This looks like a job for the Chain Rule! Differentiate the equation for the curve with respect to the parameter t. I believe you otherwise have all the information you need to answer the question.

Ok so check me if I am wrong:

for part b:

y/(2y-x) = 1/2
2y-2y= x
0 =x so the points would be (0, ± √2)

for part c:
y/(2y-x) = 0
y = 0

then 0 = 2 + x0
0 = 2 so it doesnt exist, therefore there is no point where the tangent is horizontal.

for d:

2y dy/dt = 0 + y.dx/dt + x.dy/dx
36 = 3.dx/dt + 16x
36 = 6x+3dx/dt x= 22/3

36 = 14 + 3dx/dt

22/3 = dx/dt

dynamicsolo
Homework Helper
Ok so check me if I am wrong:

for part b:

y/(2y-x) = 1/2
2y-2y= x
0 =x so the points would be (0, ± ?2)

for part c:
y/(2y-x) = 0
y = 0

then 0 = 2 + x0
0 = 2 so it doesnt exist, therefore there is no point where the tangent is horizontal.

for d:

2y dy/dt = 0 + y.dx/dt + x.dy/dx
36 = 3.dx/dt + 16x
36 = 6x+3dx/dt ; x= 22/3

36 = 14 + 3dx/dt

22/3 = dx/dt
I concur. :-)