Consider the curve given by x^ 2 +sin(xy)+3y^ 2 =C, where Cis a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01 .
dy/dx = - (y*cos(xy)+2x)/(x*cos(xy)+6y)
At (1,1), then y' = -(cos(1)+2)/(cos(1)+6) ≈ -0.388
Now, using the tangent line as an approximation, we have
y(1+∆x) = y(1) + y' ∆x = 1 + (-0.388)(.01) = 0.996
Well, if the point (1,1) lies on the curve, let's call it our "starting point". Now, we want to approximate the y-coordinate when x = 1.01, which means we're moving just a little to the right. Hmm, sounds like it's time for some tangent line action!
But before we get all tangenty, let's find the derivative of our curve with respect to x. Let me put on my math hat for a moment... ahem...
Taking the derivative of x^2 + sin(xy) + 3y^2 = C with respect to x, we get:
2x + cos(xy)(y + xy') + 6yy' = 0
Now that we have this derivative, we can plug in our starting point (x = 1, y = 1) to find the slope of the tangent line at that point. But, we don't know the value of y' (the derivative of y with respect to x) yet! So let's call it y' = m for now.
Plugging in our starting point, we get:
2(1) + cos(1*1)(1 + 1 * m) + 6(1)(m) = 0
Simplifying that lovely equation, we get:
2 + cos(1)(1 + m) + 6m = 0
Now, since we're all about approximations, let's ignore that pesky cos(1) term and pretend it's 1. So our equation becomes:
2 + 1(1 + m) + 6m = 0
And look at that, it's suddenly much simpler! Solving for m, we find:
7m = -3
m = -3/7
Alright, now we have the slope of our tangent line at the starting point. So using the point-slope form of a line, we can approximate the y-coordinate when x = 1.01.
The equation of our tangent line is:
y - 1 = (-3/7)(x - 1)
Now, let's plug in x = 1.01:
y - 1 = (-3/7)(1.01 - 1)
Simplifying that, we find:
y - 1 = (-3/7)(0.01)
y - 1 = -0.003/7
And finally, solving for y, we get:
y ≈ 1 - 0.003/7
So, the approximate y-coordinate when x = 1.01 is approximately 0.997. Voilà!
I hope my clowny approximation didn't make you feel like you're walking on a tightrope!
To approximate the y-coordinate when x = 1.01 using the tangent line approximation, we need to find the equation of the tangent line at the point (1, 1) on the curve.
Step 1: Calculate the partial derivatives of the curve equation with respect to x and y.
Let's differentiate the equation with respect to x and y:
d/dx(x^2 + sin(xy) + 3y^2) = d/dx(C)
2x + cos(xy) * (y + x * dy/dx) + 6y * dy/dx = 0 (using the product rule)
2x + cos(xy) * (y + x * dy/dx) + 6y * dy/dx = 0 (eq. 1)
d/dy(x^2 + sin(xy) + 3y^2) = d/dy(C)
sin(xy) * (x + y * dy/dy) + 6y = 0 (using the chain rule)
sin(xy) * (x + dy/dy) + 6y = 0 (eq. 2)
Step 2: Substitute the point (1, 1) into eq. 1 and eq. 2.
Let's substitute x = 1 and y = 1 into eq. 1 and eq. 2:
2(1) + cos(1 * 1) * (1 + 1 * dy/dx) + 6(1) * dy/dx = 0 (eq. 3)
sin(1 * 1) * (1 + dy/dy) + 6(1) = 0 (eq. 4)
Step 3: Solve eq. 3 for dy/dx.
Let's solve eq. 3 for dy/dx:
2 + cos(1) * (1 + dy/dx) + 6dy/dx = 0
2 + cos(1) + cos(1) * dy/dx + 6dy/dx = 0
(cos(1) + 6)dy/dx = -2 - cos(1)
dy/dx = (-2 - cos(1)) / (cos(1) + 6) (eq. 5)
Step 4: Substitute dy/dx from eq. 5 into eq. 4 and solve for dy/dy.
Let's substitute dy/dx from eq. 5 into eq. 4 and solve for dy/dy:
sin(1 * 1) * (1 + (-2 - cos(1)) / (cos(1) + 6)) + 6 = 0
sin(1) * (1 - 2 - cos(1)) + 6(cos(1) + 6) = 0
-2sin(1) - sin(1)cos(1) + 6cos(1) + 36 = 0
-2sin(1) - sin(1)cos(1) + 6cos(1) + 36 = 0
-2sin(1) + cos(1)(6 - sin(1)) + 36 = 0
cos(1)(6 - sin(1)) = 2sin(1) - 36
dy/dy = (2sin(1) - 36) / cos(1)(6 - sin(1))) (eq. 6)
Step 5: Substitute x = 1.01 into eq. 5 and eq. 6 to find the approximate y-coordinate.
Let's substitute x = 1.01 into eq. 5 and eq. 6 to find the approximate y-coordinate:
dy/dx = (-2 - cos(1)) / (cos(1) + 6)
dy/dy = (2sin(1) - 36) / cos(1)(6 - sin(1)))
After substituting the values, you can calculate the approximate y-coordinate using the tangent line approximation.
To approximate the y-coordinate when x = 1.01 using the tangent line approximation, we first need to find the equation of the tangent line at the point (1, 1) on the curve.
1. Start by finding the derivative of the curve equation with respect to x. In this case, we need to find the partial derivative ∂y/∂x, treating y as a function of x. Since ∂(sin(xy))/∂x = y*cos(xy), and ∂(3y^2)/∂x = 0, we have:
2x + y*cos(xy) + 0 = 0
2. Next, substitute the given point (1, 1) into the equation obtained in Step 1 to find the slope of the tangent line at that point. We have:
2(1) + 1*cos(1*1) = 0
2 + cos(1) = 0
cos(1) = -2
So, the slope of the tangent line at (1, 1) is -2.
3. Now, using the point-slope form of a line, we can write the equation of the tangent line as:
y - 1 = -2(x - 1)
4. Simplifying the equation, we get:
y - 1 = -2x + 2
y = -2x + 3
5. Finally, to approximate the y-coordinate when x = 1.01, substitute x = 1.01 into the tangent line equation obtained in Step 4:
y = -2(1.01) + 3
y = -2.02 + 3
y ≈ 0.98
Therefore, the y-coordinate when x = 1.01 is approximately 0.98, based on the tangent line approximation.