Calc 1

The point (1,0) lies on the curve y=sin(10π/x).
A) if Q is the point (x,sin(10π/x), find the slope of the secant line PQ.

Points are 2,1.5,1.4,1.3,1.2,1.1,0.5,0.6,0.7,0.8,0.9

Do slopes appear to be approaching a limit?

There is no 10π/x on the unit circle..... So would you put it in this format?
2, 10π/2?
For point 2 the y equals to 1.7321? How does this happen?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  1. don't use the graph of the unit circle

    instead look at the graph of the function sin((10ð/x)

    (1,0) lies on the curve because sin((10ð) = 0

    for x = 2, y = sin(5ð), which is 0 NOT 1.7321

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. you want the points

    (2, sin10π/2),(1.5, sin10π/1.5) and so on

    Do not expect the slope of the secant to approach a limit. It will oscillate between values approaching -1 and 1

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. No the answer is 1.7321 if x = 2 then y is 1.7321 that is the answer in the back

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  4. The sin function range is between -1 and +1

    1.7321 is not the sin of any angle.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  5. sin(10π/2) = sin(5π) = 0
    the slope of the line joining (1,0) and (2,0) is 0

    something is wrong with this picture.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  6. The slope of the line from the point (1.5,0.866) to the point (2,0) is -1.7321.

    But the value of the function y=sin(10Pi/x) at x=2 is y=0.

    The sine function will always return 0 when given an integer number times Pi, and 10/2 is an integer.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calc 1

    Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. y = 6 sin x, 0 ≤ x ≤ π

  2. calculus

    Find complete length of curve r=a sin^3(theta/3). I have gone thus- (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int

  3. Calculus 3

    Let r(t) = < sin(6t), cos(6t), sin(6t)cos(12t) >. Find the point where r(t) intersects the xy-plane on the interval π/6 < t < 3/12π.

  4. Mathmatics

    Find equation of tangent to curve at given point. x=cos(t)+cos(2t) y=sin(t)+sin(2t) (-1,1)

  1. AP Calculus

    Can someone check my answers: 1) Use geometry to evaluate 6 int 2 (x) dx where f(x) = { |x|, -2

  2. Maths

    Give the equation of the tangent line to the curve at given point x(t)=2 cos(t) y=2 sin(t) at t=pi/6?

  3. calculus

    A curve passes through the point (1,-11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,-16) is parallel to the x-axis. Find i) the values of a and b ii) the

  4. math;)

    Show that sin(x+pi)=-sinx. So far, I used the sum formula for sin which is sin(a+b)=sin a cos b+cos a sin b. sin(x+pi)=sin x cos pi+cos x sin pi I think I am supposed to do this next, but I am not sure. sin(x+pi)=sin x cos x+sin

  1. Calculus

    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 .

  2. math

    sin a=3/5, a lies on quadrant 1 and sin b=5/13, b lies on quadrant 2. Find cos(a+b)

  3. trig

    how do you work find sin (a-B)beta sign. sin a = 12/13,a lies in quadrant II,and cos B = 15/17, B lies in quadrant I.

  4. math

    The point P(8, 1) lies on the curve y = x − 7 . (a) If Q is the point (x, x − 7 ), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x. (i) 7.5 (ii) 7.9

View more similar questions or ask a new question.