Tangent and Normal Lines at (1,0) on Curve y = pi*sin(pi*x-y)

In summary, the slope of the tangent line is pi and the equation of the tangent line is y' = pi2 * cos(pi * x - y).
  • #1
cummings15
17
0

Homework Statement



Verify that (1,0) is on the following curve and find the tangent line and normal line to the curve at the point.

y=pisin(pix-y)

The Attempt at a Solution



i think i got it is y '
{-1/pi*cos(pi*x-y)} + pi
 
Last edited:
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  • #2
cummings15 said:

Homework Statement



Verify that (1,0) is on the following curve and find the tangent line and normal line to the curve at the point.

y=pisin(pix-y)
This is difficult to read. Use * to indicate products, like so:
y = pi * sin(pi * x - y)

cummings15 said:

The Attempt at a Solution



y prime = picos(u)(du)

y prime = picos(pix-y)(pi-y prime)

Instead of writing y prime, you can use an apostrophe - ' - to indicate a derivative.

y ' = pi * cos(pi * x - y) * (pi - y')

This is correct as far as it goes. Use algebraic techniques to get all the terms involving y' on one side and everything else on the other. Solve for y' algebraically.

Did you show that (1, 0) is a point on the graph of this equation?

Once you have solved for y', find the equation of the tangent line by evaluating y' at (1, 0). That gives you the slope of the tangent line. For the equation of the tangent line, use the point-slope form of the equation of a line.
 
  • #3
updated first post
 
Last edited:
  • #4
updated first post is y ' right
 
Last edited:
  • #5
i think i got it is y '
{-1/pi*cos(pi*x-y)} + pi
 
  • #6
cummings15 said:
i think i got it is y '
{-1/pi*cos(pi*x-y)} + pi

No, I don't think it is. Can you show us how you got that?
 
  • #7
i got this y ' = pi * cos(pi * x - y) * (pi - y')

and then i solved for y '
 
  • #8
cummings15 said:
i got this y ' = pi * cos(pi * x - y) * (pi - y')

and then i solved for y '
If you expand the right side, the equation becomes
y ' = pi2 * cos(pi * x - y) - y' * pi * cos(pi * x - y)
Get all the terms involving y' on one side, and the rest on the other side, then solve for y'.
 
  • #9
so the slope would = pi
 
  • #10
cummings15 said:
so the slope would = pi

No. Is that a guess? Why don't you substitute x=1 and y=0 into y ' = pi * cos(pi * x - y) * (pi - y') and solve for y'?
 

1. What is the tangent line?

The tangent line is a straight line that touches a curve at one point, known as the point of tangency. It represents the instantaneous rate of change or slope of the curve at that specific point.

2. How do you find the tangent line?

To find the tangent line at a specific point on a curve, you can use the slope formula, which involves finding the derivative of the function at that point. Alternatively, you can also use the geometric method of drawing a line that touches the curve at the point of interest and has the same slope as the curve at that point.

3. What is the purpose of finding the tangent line?

The tangent line is useful in many areas of mathematics and science, such as in calculus, physics, and engineering. It allows us to approximate the behavior of a curve at a specific point and make predictions about its behavior in the surrounding area.

4. Can the tangent line be found at any point on a curve?

Yes, the tangent line can be found at any point on a curve as long as the curve is continuous and differentiable at that point. In simpler terms, this means that the curve is smooth and has a defined slope at that point.

5. Are there any real-world applications of finding the tangent line?

Yes, finding the tangent line has numerous real-world applications. For example, in physics, the tangent line can be used to determine the velocity of an object at a specific point in time. In economics, the tangent line can be used to estimate the marginal cost or revenue of a product. In engineering, the tangent line can be used to design efficient and stable structures.

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