# Math Help - odd questions i'd like some advice on

1. ## odd questions i'd like some advice on

just had a quiz today and teacher threw on some questions that stumped me quite a bit and i'm trying to understand what i was supposed to do

1:
find the tangent line to the curve y = x-8/x+8 that is parallel to x-y = 8

so i differentiate the y='s equation to get (x+8) - (x-8) / (x+8)^2

that's where i'm stuck at and i don't get the parallel part

2: position of a particle is found by s = t^3 - 1.5t^2 - 2t. how long does it take to reach a velocity of 166 m/s

what i did was take the derivative of each individual term by the sum rule to get 3t^2 - 3t - 2

how do i find out when velocity reaches 166 with that equation?

2. Originally Posted by deltemis
1:
find the tangent line to the curve y = x-8/x+8 that is parallel to x-y = 8
so i differentiate the y='s equation to get (x+8) - (x-8) / (x+8)^2
The derivative is $y'=\frac{16}{(x+8)^2}$.
The slope you are looking for is $1$ and that happens if $x=-4$.

3. hmm i differentiated wrong. i'm wondering how you found the slope and at what x it occurs?

also does anybody know how to get my second problem? i'm having no luck with whatever i try

4. Originally Posted by deltemis
[snip]

2: position of a particle is found by s = t^3 - 1.5t^2 - 2t. how long does it take to reach a velocity of 166 m/s

what i did was take the derivative of each individual term by the sum rule to get 3t^2 - 3t - 2

how do i find out when velocity reaches 166 with that equation?
ds/dt = v = 3t^2 - 3t - 2. Substitute v = 166 and solve for t.

Originally Posted by deltemis
hmm i differentiated wrong. i'm wondering how you found the slope and at what x it occurs?[snip]
x-y = 8 is a line and has a gradient of 1. Tangent parallel to x-y = 8 means gradient of tangent = 1. Substitute y' = 1 and solve for x.