# Thread: Tangent Lines and derivatives at a point

1. ## Tangent Lines and derivatives at a point

This is a very simple problem but I can't seem to get the correct answer for some reason.

Question: Find the slope of the function's graph at the given point.

h(t) = t^3 + 6t, (3, 45). Also. Find an equation for the line tangent to the graph there.

My answer is m=21 for the slope and for the equation of the tangent line, y=21x-54. Maybe I'm approaching the problem incorrectly?

2. Originally Posted by casemeister
This is a very simple problem but I can't seem to get the correct answer for some reason.

Question: Find the slope of the function's graph at the given point.

h(t) = t3 + 6t, (3, 45). Also. Find an equation for the line tangent to the graph there.

My answer is m=21 for the slope and for the equation of the tangent line, y=21x-54. Maybe I'm approaching the problem incorrectly?
If you post your working, it'll be easy to point out your mistake. In fact, I'm betting that if you do this you'll actually spot the mistake yourself .......

3. Originally Posted by mr fantastic
In fact, I'm betting that if you do this you'll actually spot the mistake yourself .......
I highly doubt it, but here it is anyways.

t^3-6t, Differentiates into f '(x) = 3t^2+6 . Factors into: 3(t^2+2)then I substituted 3 into t.
f(3)=3^3+6 = 33 . then i used the point slope formula to find the tangent line.

y-9 = 33(x-3) = y=33x-90

Ok, I got the slope correct, but my tanget line is still incorrect.

4. Originally Posted by casemeister
I highly doubt it, but here it is anyways.

t^3-6t, Differentiates into f '(x) = 3t^2+6 . Factors into: 3(t^2+2)then I substituted 3 into t.
f(3)=3^3+6 = 33 . then i used the point slope formula to find the tangent line.

y-9 = 33(x-3) = y=33x-90

Ok, I got the slope correct, but my tanget line is still incorrect.
Using the formula for the equation of a tangent line

$y-h(x_0)=h'(x_0)(x-x_0)$
with $h(x_0)=45$

$x_0=3$

and $h'(x)=3x^2+6\Rightarrow{h'(3)=3(3)^2+6=33}$

Imputting we get

$y-45=33(x-3)$

Can you go from there?

5. Originally Posted by Mathstud28
Using the formula for the equation of a tangent line

$y-h(x_0)=h'(x_0)(x-x_0)$
with $h(x_0)=45$

$x_0=3$

and $h'(x)=3x^2+6\Rightarrow{h'(3)=3(3)^2+6=33}$

Imputting we get

$y-45=33(x-3)$

Can you go from there?
I hope so,

$y-45=33(x-3)$

$y=33x-54$

thats what i got, but its still incorrect.

6. Originally Posted by casemeister
I hope so,

$y-45=33(x-3)$

$y=33x-54$

thats what i got, but its still incorrect.
What makes you think that this is incorrect?

7. thats what i screamed at the computer because apparently it isn't the correct answer on the class homework web program. maybe its my formatting but even then its still not taking it.

8. Originally Posted by casemeister
thats what i screamed at the computer because apparently it isn't the correct answer on the class homework web program. maybe its my formatting but even then its still not taking it.
Unless you have misposted the question then I am pretty sure the computer is wrong. I assure you the method I showed is sound in mathematical validity. Don't stress out =)...you got the right answer....you even beat the computer

9. lol, i figured it out. since the question has it in terms of t, i expressed my answer in x and it wouldnt take it, go figure. computers...right?

10. Originally Posted by casemeister
lol, i figured it out. since the question has it in terms of t, i expressed my answer in x and it wouldnt take it, go figure. computers...right?
...crazy computers

11. Originally Posted by mr fantastic
[snip]
In fact, I'm betting that if you do this you'll actually spot the mistake yourself .......
Originally Posted by casemeister
I highly doubt it, but here it is anyways.

t^3-6t, Differentiates into f '(x) = 3t^2+6 . Factors into: 3(t^2+2)then I substituted 3 into t.
f(3)=3^3+6 = 33 . then i used the point slope formula to find the tangent line.

y-9 = 33(x-3) = y=33x-90

Ok, I got the slope correct, but my tanget line is still incorrect.
So I win the bet. Especially since you eventually figured out that you needed t not x as the variable.

Be honest now .....Wasn't this whole process more satisfying than someone typing out the solution right off the bat .....?