# Thread: Finding an equation that is tangent to a curve and whose slope is a max.

1. ## Finding an equation that is tangent to a curve and whose slope is a max.

a) Find an equation of a line that is tangent to the curve f(x)=2cos3x and whose slope is a maximum.

b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?

2. Originally Posted by kmjt
a) Find an equation of a line that is tangent to the curve f(x)=2cos3x and whose slope is a maximum.

b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
First derivative test.

3. Differentiation once gives the slope of the function .

So differentiation after differentiation , you would find the extrema of the slope !

Differentiation thrice you would test whether it is max .

4. Hmm could I possibly have an example? I've never done the first derivative test with a sinusoidal function.. only basic stuff like f(x)=x^3-6x^2+15 for example

5. Originally Posted by kmjt
Hmm could I possibly have an example? I've never done the first derivative test with a sinusoidal function.. only basic stuff like f(x)=x^3-6x^2+15 for example
Suppose f(x)=sinx

$\displaystyle f'(x)=cos(x)=0$

Where does $\displaystyle cos(x)=0$?

$\displaystyle \frac{\pi}{2}+\pi k$

Thus your critical values are at those points.

Then you check the left and right of the values to determine if it is a relative max or min.

Since this function is wave, all the rel. max will be the same.

6. Originally Posted by kmjt
For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
Simple. Just take the derivative of that again (i.e. the second derivative). So the original function is

$\displaystyle y = 2cos(3x)$

then..

$\displaystyle \frac{dy}{dx} = -6cos(3x)$

to find when the derivative is at a maximum, we take the derivative once more

$\displaystyle \frac{d^2y}{dx^2} = -18sin(3x)$

and we find that it's at a maximum when $\displaystyle x = \frac{\pi}{6}$.

So than we evaluate the first derivative to see what the slope is at $\displaystyle \frac{\pi}{6}$.

$\displaystyle \frac{dy(\frac{\pi}{6})}{dx} = -6$

so the slope, which happens to be the optimal slope, is -6.

To find where on the graph we should draw the tangent line, we evaluate $\displaystyle \frac{\pi}{6}$ using the regular function and we find that

$\displaystyle y(\frac{\pi}{6}) = 0$

using this info we can do some algebra to get the equation of this line which is

$\displaystyle y = -6x + \pi$

b) This isn't the only possible solution. Remember when we took the second derivative

$\displaystyle \frac{d^2y}{dx^2} = -18sin(3x)$

and set it equal to 0 and got $\displaystyle x = \frac{\pi}{6}$. Then we evaluated $\displaystyle \frac{\pi}{6}$ using the original function, $\displaystyle y = 2cos(3x)$, and we got $\displaystyle y(\frac{\pi}{6}) = 0$.

Essentially what this is telling us is that the original function hits a slope of $\displaystyle \pm6$ when ever it touches the x-axis (equal to 0).

Since it's a trig function, it continues to hit 0 to infinity.

The real answer (when the function hits a maximum slope) is $\displaystyle x = \frac{\pi}{6} + k\frac{\pi}{3}$ where $\displaystyle k$ is an integer.

Cheers!

7. Originally Posted by eddie2042

to find when the derivative is at a maximum, we take the derivative once more

$\displaystyle \frac{d^2y}{dx^2} = -18sin(3x)$

and we find that it's at a maximum when $\displaystyle x = \frac{\pi}{6}$.
How do you know the max is at $\displaystyle x = \frac{\pi}{6}$ from looking at -18sin(3x)?

8. Originally Posted by kmjt
How do you know the max is at $\displaystyle x = \frac{\pi}{6}$ from looking at -18sin(3x)?

uh-oh.. i just notcied a mistake. in my original post i said the derivative is

$\displaystyle \frac{dy}{dx} = -6cos(3x)$ <<<< WRONG!

it's supposed to be

$\displaystyle \frac{dy}{dx} = -6sin(3x)$ <<<<< CORRECT!

this means that the second derivative is also wrong. it was

$\displaystyle \frac{d^2y}{dx^2} = -18sin(3x)$ <<<< WRONG!

it's supposed to be

$\displaystyle \frac{d^2y}{dx^2} = -18cos(3x)$<< CORRECT!

ok now we're back on track. We set the second derivative equal to zero, and we find the slope is at a max when

$\displaystyle -18cos(3x) = 0$

$\displaystyle 3x = arcos(0)$

$\displaystyle x = \frac{\pi}{6}$

Everything's correct from there.. i mention the second derivative again; replace it with the corrected one...

9. $\displaystyle 3x = arcos(0)$

I'm not quite sure what that means. I understand everything else.

10. Originally Posted by kmjt
$\displaystyle 3x = arcos(0)$

I'm not quite sure what that means. I understand everything else.
I'm trying to solve for x. We started with the equation

$\displaystyle -18cos(3x) = 0$

then we divide both sides by -18

$\displaystyle cos(3x) = 0$

then we take the inverse cos function of both sides (sometimes it's written with a power of minus one but i prefer $\displaystyle arcos$. so..

$\displaystyle 3x = arccos(0)$
$\displaystyle 3x = \frac{\pi}{2}$ .. inverse cos of 0 is 90 degrees (or pi/6 radians)

and finally...

$\displaystyle x = \frac{\pi}{6}$

11. Originally Posted by kmjt
a) Find an equation of a line that is tangent to the curve f(x)=2cos3x and whose slope is a maximum.

b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
this is a situation where your knowledge of trig graphs goes a long way in determining the maximum slope with little calculus manipulation.

$\displaystyle y = 2\cos(3x)$ has a period of $\displaystyle T = \frac{2\pi}{3}$, and given the general shape of the cosine curve, the maximum slope will occur when $\displaystyle 3x = \frac{3\pi}{2}$ ... $\displaystyle x = \frac{\pi}{2}$.

this maximum slope will occur at all values of $\displaystyle x = \frac{\pi}{2} + k \cdot \frac{2\pi}{3} \, ; \, k \in \mathbb{Z}$ , and will be equal to 6.

12. $\displaystyle x = \frac{\pi}{2} + k \cdot \frac{2\pi}{3} \, ; \, k \in \mathbb{Z}$

What does that mean? I just sketched the graph of 2cos3x and I see how there is a max at every 2pi/3 interval.

13. Originally Posted by kmjt
$\displaystyle x = \frac{\pi}{2} + k \cdot \frac{2\pi}{3} \, ; \, k \in \mathbb{Z}$

What does that mean? I just sketched the graph of 2cos3x and I see how there is a max at every 2pi/3 interval.
It means that at all those points of x slope is max, for a k (constant integer) within the region Z

14. Originally Posted by AllanCuz
It means that at all those points of x slope is max, for a k (constant integer) within the region Z
region Z ?

$\displaystyle k \in \mathbb{Z}$ means $\displaystyle k$ is an element of the set of integers ... $\displaystyle \mathbb{Z}$ is the symbol for the set of integers.

15. Originally Posted by skeeter
region Z ?

$\displaystyle k \in \mathbb{Z}$ means $\displaystyle k$ is an element of the set of integers ... $\displaystyle \mathbb{Z}$ is the symbol for the set of integers.
Yeah, my bad on that one..good thing I'm not a prof yikes

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