equations for tangent lines?

Question

1) Find an equation for the tangent line in x and y.
x(t) = 2t - 1
y(t) = t^4
t = 1

2)Find an equation for the tangent line to the following polar curve in x and y.
r = 1/[2-sin(theta)]
theta = pi

this is NOT my homework or anything like that...

Answer 1

x and y are both functions, to find the tangent line u can rewrite the equations

x(t) = 2t - 1
y(t) = t^4

x(t) = 2t - 1 ==> t= (x + 1)/2

y(t) = t^4 ==> y = [x/2 + 1/2]^4

take the derivative dy/dx

dy/dx = (1/2)*4*[x/2 + 1/2]^3

for t=1, x=2(1)-1=1, y=1^4=1

for t=1, dy/dx = 2 (check the calculations) so m=2

y -1 = 2(x - 1) ==> y = 2x -1

use ur calculator and graph it to check the answer.

there is anoterh way to find the derivative:

dy/dx = (dy/dt)/(dx/dt) = (4t^3)/(2) = 2t^3 = 2[x/2 + 1/2]^3

this is a better way, if u are samiliar with. but u get the same answer in both ways.

2)

Answer 2

I take it you already comprehend derivatives. f(x)= (x^3 - x^2 + x)^8so discover f '(x) f '(x)=8(x^3 - x^2 + x)^7*(3x^2-2x+a million) it really is through the chain rule g(x)^8=8*g(x)^7*g '(x) now you want the slope at the same time as x=a million so f '(a million)=16 now you want a line through (a million,a million) and slope 16 so we use our factor slope sort with m=16 and factor (a million,a million) y – y1 = m(x – x1) and x1=y1=a million so y-a million=16(x-a million) or y-a million=16x-16 or y=16x-15 is the line through a million,a million and tangent to the given f(x).

Answer 3

dx = 2 dt; dy = 4t dt
dy/dx = 2/(4t) = 1/(2t)
For t=1, x(1) = 1; y(1) = 1; dy/dx = 1/2

Your tangent line passes through (1,1) with a slope of 1/2. Using point-slope,

(y-1)/(x-1) = 1/2
2y - 2 = x - 1
x - 2y = -1 (answer in standard form)

r = 1/(2 - sin t) (where t is theta)
r = (2 - sin t)^(-1)
dr/dt = (-1)(2 - sin t)^(-2)(-cos t)
dr/dt = (cos t)/(1 - 2 sin t)^2

We have x = r cos t, y = r sin t
dx/dt = r(- sin t) + cos t (dr/dt) = (dr/dt) cos t - r sin t
dy/dt = r cos t + sin t (dr/dt) = (dr/dt) sin t + r cos t

This looks like a mess, but maybe we can avoid part of it by plugging in values.

dy/dx = (dy/dt)/(dx/dt)

Here, let's evaluate for t = pi:
r(pi) = 1/2
dr/dt = (cos pi)/(1 - 2 sin pi)^2 = -1 (at t = pi)
x(pi) = r cos pi = -r = -1/2
y(pi) = r sin pi = 0 (that makes life easy!)
dx/dt = (dr/dt) cos t - r sin t = (-1)(-1) - (1/2)(0) = 1
dy/dt = (dr/dt) sin t + r cos t = (-1)(0) + (1/2)(-1) = -1/2
dy/dx = (dy/dt)/(dx/dt) = (-1/2)/1 = -1/2

So we need the line with slope (-1/2) passing through (-1/2, 0) (these last are the x,y coordinates for t = pi). Using point-slope:

(y - 0)/(x + 1/2) = -1/2
2y = -x - 1/2
4y = -2x - 1
2x + 4y = -1

That's the answer in standard form. (Unless, of course, I made a mistake. Please check my work.)