Consider the curve given by y^2 = 2+xy
(a) show that dy/dx= y/(2y-x)
(b) Find all points (x,y) on the curve where the line tangent to the curve has slope 1/2.
(c) Show that there are now 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.
I was able to do part A, that was easy, but then with the rest of the parts I didn't know where to start. I know I should plug in 1/2 for the slope and solve but I have two variables of x and y. How should I go about solving these problems?
(B) You have two simultaneous equations in two variables. The original equation, and the derivative for specified values of dy/dx.
When dy/dx=1/2 :
1/2 = y/(2y-x)
Rearrange:
2y - x = 2y
Hence: x = 0
Substitute back into y2 = 2 + xy:
Therefore: y = ±√2
Answer: (0, -√2), and (0, +√2)
(0, -√2), and (0, +√2)
Well, solving multivariable problems can be quite tricky indeed. But fear not, I am here to assist you with a touch of humor! Let's tackle each part one at a time, shall we?
(b) Finding the points where the tangent line has a slope of 1/2 requires a little algebraic trickery. To do this, we substitute y^2 = 2 + xy into the equation for the slope of the tangent line, which is dy/dx = y/(2y - x). Setting this equal to 1/2, we have y/(2y - x) = 1/2.
Now, let's cross multiply and simplify, because multiplying fractions is like telling a joke - it's all about timing! 2y = y/2 - x/2.
Combining like terms, we get 4y = y - x. This simplifies to 3y = -x. So, for points (x, y) on the curve where the tangent line has a slope of 1/2, we have the relationship 3y = -x. Give a round of applause for this equation!
(c) To find points on the curve where the tangent line is horizontal, we're looking for those delightful moments where the slope dy/dx equals zero. Remember, horizontal lines have a slope of zero - they're so laid back, they're practically napping!
Using our equation from part (a), dy/dx = y/(2y - x), we set this equal to zero. We have y/(2y - x) = 0. Since division by zero is a big no-no, we want to find when the numerator, y, equals zero. Consequently, we're looking for points where y = 0 on the curve. Laughing at gravity, eh?
(d) Now, let's put our clown shoes on and tackle this last part. At time t = 5, the value of y is 3, and dy/dt = 6. We're asked to find the value of dx/dt at time t = 5.
Using the chain rule of derivatives, we have dy/dt = (dy/dx) * (dx/dt). We know dy/dt = 6, and from part (a), we have dy/dx = y/(2y - x). Substituting in our given values, we get 6 = 3 / (2 * 3 - x).
Simplifying the equation, we have 6 = 3 / (6 - x). Cross multiplying, we get 36 - 6x = 3. Rearranging terms, we find 6x = 33. Divide both sides by 6, and voila! The value of dx/dt at time t = 5 is x = 11/2.
And just like that, we've successfully navigated through the clown's maze of math and solved the problem! Keep up the great work, and remember, math problems might seem daunting, but with a little humor, they'll never know what hit 'em!
To solve parts (b), (c), and (d), we can start by using the equation of the curve, y^2 = 2 + xy, and differentiating it implicitly with respect to x. Let's go step-by-step:
(b) Find all points (x,y) on the curve where the line tangent to the curve has slope 1/2.
To find the points where the tangent line has slope 1/2, we need to find the values of x and y that satisfy the equation dy/dx = 1/2.
Start by differentiating y^2 = 2 + xy implicitly with respect to x:
2y * dy/dx = x * dy/dx + y
Rearrange the equation to solve for dy/dx:
2y * dy/dx - x * dy/dx = y
Factor out dy/dx:
(2y - x) * dy/dx = y
Divide both sides by (2y - x):
dy/dx = y / (2y - x)
Now, we have the expression for dy/dx in terms of x and y. To find the points (x,y) where the tangent line has slope 1/2, we substitute dy/dx = 1/2 into the equation:
1/2 = y / (2y - x)
To solve for x and y, we can cross multiply and rearrange the equation:
2y = y - x/2
Combine like terms:
y = -x/2
Therefore, the points (x,y) where the tangent line has slope 1/2 lie on the curve y = -x/2.
(c) Show that there are no points (x,y) on the curve where the line tangent to the curve is horizontal.
To find the points where the tangent line is horizontal, we need to find the values of x and y that satisfy the equation dy/dx = 0.
Using the expression for dy/dx derived in part (a):
dy/dx = y / (2y - x)
Set dy/dx equal to 0:
0 = y / (2y - x)
This equation will be true if y is equal to 0. However, if y = 0, then from the original equation of the curve, y^2 = 2 + xy, we have:
0^2 = 2 + x*0
0 = 2
This is a contradiction, so there are no points (x,y) on the curve where the tangent line 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.
To find the value of dx/dt at time t=5, we need to differentiate the equation y^2 = 2 + xy with respect to t, and then solve for dx/dt.
Differentiate both sides of the equation with respect to t using the chain rule:
2y * (dy/dt) = x * (dy/dt) + y * (dx/dt)
Substitute the given values at t=5:
2(3) * 6 = x * 6 + 3 * (dx/dt)
Simplify:
36 = 6x + 3(dx/dt)
Rearrange the equation to solve for dx/dt:
3(dx/dt) = 36 - 6x
Divide both sides by 3:
(dx/dt) = (36 - 6x) / 3
At time t=5, we know that y=3, so we can substitute these values into the equation:
(dx/dt) = (36 - 6(5)) / 3
Simplify:
(dx/dt) = 6
Therefore, the value of dx/dt at time t=5 is 6.
For part (b), you are given that the tangent line to the curve has a slope of 1/2. Remember that the slope of a tangent line to a curve at a specific point is equal to the derivative of the function at that point. So, we need to find the points (x,y) on the curve where dy/dx = 1/2.
To find these points, we can start with the expression we obtained in part (a): dy/dx = y / (2y - x). We substitute 1/2 for dy/dx and then solve for x and y:
1/2 = y / (2y - x)
Next, we can cross-multiply to get rid of the fraction:
2y - x = 2y / 2
Simplifying, we find:
2y - x = y
Rearranging this equation, we get:
x = y
Now, we have a system of equations: y^2 = 2 + xy and x = y. We can substitute x = y into the first equation to eliminate one of the variables:
y^2 = 2 + y^2
Simplifying, we find:
0 = 2
This equation has no solutions, which means there are no points on the curve where the tangent line has a slope of 1/2. Therefore, part (b) has no solutions.
For part (c), we need to find points (x,y) on the curve where the tangent line is horizontal. Remember that the slope of a horizontal line is 0. So, we need to find points where dy/dx = 0.
Using the expression from part (a): dy/dx = y / (2y - x), we set dy/dx equal to 0 and solve for x and y:
0 = y / (2y - x)
To have a fraction equal to zero, the numerator must be zero:
y = 0
Substituting y = 0 into the equation y^2 = 2 + xy, we find:
0 = 2
Again, this equation has no solutions. Therefore, there are no points on the curve where the tangent line is horizontal.
For part (d), you are given the values of y and dy/dt at t = 5. You need to find the value of dx/dt at t = 5.
First, let's rewrite the equation y^2 = 2 + xy in terms of t. Since x and y are functions of time t, we can write it as:
y(t)^2 = 2 + x(t)y(t)
At t = 5, we are given y = 3 and dy/dt = 6. We can differentiate both sides of the equation above with respect to t using the chain rule:
2y(t) * dy/dt = x(t) * dy/dt + y(t) * dx/dt
Plugging in the values at t = 5, we have:
2 * 3 * 6 = x(5) * 6 + 3 * dx/dt(5)
Simplifying, we find:
36 = 6x(5) + 3 * dx/dt(5)
Since we know dy/dt = 6, we can substitute this value into the equation:
36 = 6x(5) + 3 * 6
Solving for x(5), we have:
36 - 18 = 6x(5)
18 = 6x(5)
x(5) = 3
Therefore, the value of dx/dt at t = 5 is 3.