Could someone please help me with these tangent line problems?
1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1.
2) Show that there is no point on the graph of x^2 - 3xy + y^2 = 1 where the line tangent to the curve is horizontal.
3) Find all points on the curve xy^2 - x^3y = 6 where the tangent line is vertical. Explain your reasoning.
3y^3 + 2x^2 = 5
9y^2 dy/dx + 4x = 0
dy/dx = -4x/9y^2
when y = 1 , 3 + 2x^2 = 5
2x^2 = 2
x = ± 1 , but in quadrant 1, x = 1
dy/dx = -4(1)/9 = -4/9
equation of tangent:
y-1 = (-4/9)(x-1)
9y-9 = -4x + 4
4x + 9y = 13
check:
http://www.wolframalpha.com/input/?i=plot+3y%5E3+%2B+2x%5E2+%3D+5,+4x+%2B+9y+%3D+13
notice the tangent at (1,1)
2.
x^2 - 3xy + y^2 = 1
2x -3x(dy/dx) - 3y + 2y dy/dx = 0
dy/dx(-3x + 2y) = 3y - 2x
dy/dx = (3y - 2x)/(2y - 3x)
to have a horizontal slope, dy/dx = 0 , that is,
3y - 2x = 0
y = 2x/3
sub that back into the original
x^2 - 3x(2x/3) + 4x^2/9 = 1
times 9
9x^2 - 18x^2 + 4x^2 = 1
-5x^2 = 1
There is no real solution to this, so there is no such tangent.
3.
find the derivative like I showed you in the first two.
Since the tangent is vertical, the denominator of the slope must be zero.
Show that this is not possible.
Thank you so much! And I will
Sure, I can help you with these tangent line problems. Here's a step-by-step explanation on how to solve each problem:
1) To find the equation of the tangent line to the curve 3y^3 + 2x^2 = 5 at a point where y = 1, you need to find the derivative of the function and evaluate it at that point.
- Take the derivative of the equation with respect to x, treating y as a constant. This will give you the slope of the tangent line.
- The derivative of 3y^3 + 2x^2 with respect to x is 4x.
- Plug in the values of x and y from the point of tangency (in this case, y = 1) into the derivative. This will give you the slope of the tangent line at that point.
- Substitute x = x-coordinate of the point of tangency and y = y-coordinate of the point of tangency into the original equation. This will give you an equation with x only which represents the line tangent to the curve at that point.
- Therefore, the equation of the tangent line is obtained.
2) To show that there is no point on the graph of x^2 - 3xy + y^2 = 1 where the line tangent to the curve is horizontal, you can use the same reasoning.
- Take the derivative of the equation with respect to x. This will give you the slope of the tangent line.
- Find the critical points by setting the derivative equal to zero and solving for x and y.
- If there are no real solutions for x and y, then there are no points on the graph where the tangent line is horizontal.
3) To find all points on the curve xy^2 - x^3y = 6 where the tangent line is vertical, you need to find the derivative with respect to x and analyze the behavior of the derivative.
- Take the derivative of the equation with respect to x. This will give you the slope of the tangent line.
- Set the derivative equal to infinity (or undefined) to find the x-values where the tangent line is vertical.
- Solve the resulting equation to find the x-values.
- Substitute these x-values back into the original equation to find the corresponding y-values.
- Therefore, the points where the tangent line is vertical are obtained.
Remember, finding the derivative and solving equations are key steps in solving these types of tangent line problems. By following these steps, you should be able to find the equations of the tangent lines and determine the points of tangency for each problem.