Math Help - Fine Tangent Line

1. Fine Tangent Line

Hi! I am new to this forum I am looking forward to discuss math problems with everyone here ;]

1)a. Find the tangent line to the curve f(x) = 2^sin(x) at the point x = 0

my answer is y = 1+ln2[2^sin(x)]cos(x)[x]

I don't know what I did wrong

b. Use the line in part (a) to estimate the value of f(.5)
my value = 1.424 I got it wrong also ><

I used this formula: f(x) = f(a) + f'(a)(x-a)

c. What is the error in the approximation, accurate to three decimal places?
my answer is -.454 and it's also wrong ><

I used this formula: E(x) = F(x) - L(x)

can someone please explain to me what I did wrong???
I can't figure it out

2. Originally Posted by Ryuu
1)a. Find the tangent line to the curve f(x) = 2^sin(x) at the point x = 0

my answer is y = 1+ln2[2^sin(x)]cos(x)[x]

I don't know what I did wrong
$f(x) = 2^{sin(x)}$

Thus
$f^{\prime}(x) = ln(2) \cdot cos(x)~2^{sin(x)}$

At x = 0:
$f^{\prime}(0) = ln(2) \cdot cos(0)~2^{sin(0)} = ln(2)$

So you are looking for a line with a slope of ln(2) that passes through the point (0, 1).
$y = mx + b$

$1 = ln(2) \cdot 0 + b$

Thus
$y = ln(2)x + 1$

-Dan

3. Originally Posted by topsquark
$f(x) = 2^{sin(x)}$

Thus
$f^{\prime}(x) = ln(2) \cdot cos(x)~2^{sin(x)}$

At x = 0:
$f^{\prime}(0) = ln(2) \cdot cos(0)~2^{sin(0)} = ln(2)$

So you are looking for a line with a slope of ln(2) that passes through the point (0, 1).
$y = mx + b$

$1 = ln(2) \cdot 0 + b$

Thus
$y = ln(2)x + 1$

-Dan

How did you get point (0,1) can't it be (0,0) ??
other than that thank you for the explanation

OH OKAY you plug in the 0 to the original equation and get 1

okay okay THANK YOU =]

now part B and C I still need help ><

4. Originally Posted by Ryuu
Hi! I am new to this forum I am looking forward to discuss math problems with everyone here ;]

1)a. Find the tangent line to the curve f(x) = 2^sin(x) at the point x = 0

my answer is y = 1+ln2[2^sin(x)]cos(x)[x]

I don't know what I did wrong

b. Use the line in part (a) to estimate the value of f(.5)
my value = 1.424 I got it wrong also ><

I used this formula: f(x) = f(a) + f'(a)(x-a)

c. What is the error in the approximation, accurate to three decimal places?
my answer is -.454 and it's also wrong ><

I used this formula: E(x) = F(x) - L(x)

can someone please explain to me what I did wrong???
I can't figure it out

Hello,

to a) your calculations are OK - you only mixed the calculation of the slope with the claculation of the equation of the tangent:

the slope of the tangent is $m =\ln(2) \cdot (2^{\sin(0)}) \cdot \cos(0) = \ln(2) \cdot 2^0 \cdot 1 = \ln(2)$

Thus the equation of the tangent is:

$t:\ y=\ln(2) \cdot x + 1$

to b) Plug in x = 0.5 into the equation of the tangent (that means you use the tangent line as an approximation of the function):

$f(0.5) \approx \ln(2) \cdot 0.5 + 1 \approx 1.34657...$

to c) I'm not certain what you should do here. The nearly exact value of

$f(0.5) = 1.394188409$

5. Originally Posted by earboth
Hello,

to a) your calculations are OK - you only mixed the calculation of the slope with the claculation of the equation of the tangent:

the slope of the tangent is $m =\ln(2) \cdot (2^{\sin(0)}) \cdot \cos(0) = \ln(2) \cdot 2^0 \cdot 1 = \ln(2)$

Thus the equation of the tangent is:

$t:\ y=\ln(2) \cdot x + 1$

to b) Plug in x = 0.5 into the equation of the tangent (that means you use the tangent line as an approximation of the function):

$f(0.5) \approx \ln(2) \cdot 0.5 + 1 \approx 1.34657...$

to c) I'm not certain what you should do here. The nearly exact value of

$f(0.5) = 1.394188409$

Okay! I got part B now, which mean i don't need to use the formula ><
I still don't get part C ><

how did you get that exact value? and doesn't mean you have to subtract to get the error in approximation?

Thanks