hey can someone help me with these i need help

Consider the curve given by x2 + sin(xy) + 3y2 = C, where C is a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01.

Question 19 options:

0.996


1


1.004


Cannot be determined


1.388

The volume of an open-topped box with a square base is 245 cubic centimeters. Find the height, in centimeters, of the box that uses the least amount of material.


7.883 centimeters


6 centimeters


3.942 centimeters


3 centimeters


2 centimeters

x2 + sin(xy) + 3y2 = C

solve for dy/dx
2x+cos(xy)xdy/dx+6ydy/dx=0

dy = -2x/(cos(xy)x+6y) * dx

dy= -2*1/(cos(1*1) + 6) dx
but dx= .001
then dy= -.02/(.540+6)= -0.00305810398
so y= 1-dy= 0.996941896

Sure, I can help you with these questions. Let's start with the first one.

To approximate the y-coordinate when x = 1.01, we need to use the tangent line approximation. Here's how you can do it:

1. First, find the derivative of the given curve with respect to x. The derivative will give you the slope of the tangent line at any given point on the curve.

2. Once you have the derivative, evaluate it at the point (1, 1) to find the slope of the tangent line at that point.

3. Use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point on the curve (in this case, (1, 1)) and m is the slope of the tangent line.

4. Plug in the value x = 1.01 into the equation found in step 3 and solve for y to find the approximate y-coordinate.

Now let's move to the second question.

To find the height of the box that uses the least amount of material, we need to optimize the box's dimensions. Here's how to do it:

1. Let's assume the side length of the square base of the box is s centimeters, and the height of the box is h centimeters. Since the volume of the box is 245 cubic centimeters, we can write an equation relating s and h: s^2 * h = 245.

2. We can rewrite the equation from step 1 as h = 245 / s^2.

3. Now we need to minimize the amount of material used, which is the surface area of the box. The surface area consists of the area of the square base (s^2) and the area of the four vertical sides (4sh).

4. Substitute the equation from step 2 into the surface area equation: surface area = s^2 + 4sh.

5. Differentiate the surface area equation with respect to s to find the critical points. Set the derivative equal to zero and solve for s.

6. Once you find the critical value(s) of s, substitute it back into the equation from step 2 to find the corresponding height h.

By following the steps above, you should be able to find the approximate y-coordinate for the first question and the height of the box for the second question. Good luck!

Sure, I can help you with these problems!

For the first question:
1. To use the tangent line approximation, we need to find the partial derivatives of the curve equation with respect to x and y.
2. Taking the partial derivative of the given equation with respect to x, we get:

(d/dx)(x^2 + sin(xy) + 3y^2) = 2x + y*cos(xy)

3. Evaluate the above expression using the coordinates of the given point (1, 1):

2(1) + 1*cos(1*1) = 2 + cos(1)

4. Now, we can find the slope of the tangent line at this point.
5. Using the formula for a linear approximation, the equation of the tangent line is given by:

y - y1 = m(x - x1), where (x1, y1) are the coordinates of the given point.

6. Substituting the values, we have:

y - 1 = (2 + cos(1))(x - 1)

7. Substitute x = 1.01 into the equation and solve for y to get the approximate value of the y-coordinate.

For the second question:
1. Let's assume the side length of the square base is x centimeters, and the height of the box is h centimeters.
2. The equation for the volume of the box is given by: V = x^2 * h = 245.
3. We need to find the value of h that minimizes the surface area (amount of material used) of the box.
4. The surface area of the box is given by: A = x^2 + 4xh.
5. Rewrite the volume equation in terms of h: h = 245 / x^2.
6. Substitute the value of h in the surface area equation: A = x^2 + 4x(245 / x^2).
7. Simplify the equation to A = x^2 + 980 / x.
8. Take the derivative of A with respect to x and set it equal to zero to find the critical points.
9. Solve for x to find the side length of the base of the box that generates the minimum surface area.
10. Substitute the value of x into h = 245 / x^2 to find the height in centimeters.