English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Find the equation of the tangent line to the curve at the point corresponding to the given value of the parameter:

x = t^4 + 1
y = t^3 + t
t = -1

The answer is -x, I just don't know how to go about solving it...

m (slope) = y'/x' (y-prime divided by x-prime) = (3t^2 + 1)/(4t^3) | t = -1; m = -1?

So I guess:
x = (-1)^4 + 1 = 2
y = (1)^3 + (-1) = 0

y - y1 = m(x - x1)
y - 0 = -1(x - 2)
y = -x + 2

So I don't know...either the book is wrong, or I'm wrong. I'm betting I'm wrong :)

2006-10-17 11:19:41 · 4 answers · asked by Derek 4 in Science & Mathematics Mathematics

4 answers

y = (1)^3 + (-1) = 0
.......↑this is the source of your problem. It should be -1, which gives you y=-2. Then in your point-slope formula, you would have y+2=-1(x-2) → y+2=-x+2 → y=-x

2006-10-17 11:29:13 · answer #1 · answered by Pascal 7 · 0 1

your y value is miscalculated
y=t^3 +t when t= -1 gives
y=(-1)^3 + (-1)= -1 +-1 = -2
that should fix your problem.
:)

2006-10-17 11:30:51 · answer #2 · answered by mikedotcom 5 · 0 0

Check your y-value again. It should be y=(-1)^3+(-1)=-2.

2006-10-17 11:28:48 · answer #3 · answered by mathematician 7 · 1 0

you're given 2 equations, x=f(t) and y=g(t). get mutually: x=2t and y=t+a million, for t in [-2, 3] you plug in -2 and three in for t in the x equation. this promises a sparkling area of [-4, 6]. to sparkling up for oblong variety: sparkling up the x equation for t. subsequently, x/2=t. Plug in x/2 for the t in the y equation. Your oblong variety is y=x/2+a million.

2016-12-04 22:40:51 · answer #4 · answered by Anonymous · 0 0

fedest.com, questions and answers