Answer to Question #240784 in Calculus for Sakshi

Question

a) Find the linear and quadratic approximations of

sin(0.96π) tan(0.26π) + (0.96)^2 (0.26)

b) Compare these approximations with the actual value.

Answer 1

(a) Given function is: $f(x)=sin(0.96π) tan(0.26π) + (0.96)^2 (0.26)$

Since no point is given so we are calculating the value at point 0.

The linear approximation to $f(x)=\sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625} \ at \ x_{0}=0 \\$

A linear approximation is given by: $L(x) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) .$

We are given that $x_{0}=0 .$

Firstly, find the value of the function at the given point:

$y_{0}=f\left(x_{0}\right)=\sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625}$

Secondly, find the derivative of the function, evaluated at the point: $f^{\prime}(0) .$

Next, evaluate the derivative at the given point to find slope.

$f^{\prime}(0)=0$

Plugging the values found, we get that: $L(x) \approx \sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625}+0(x-(0)) .$

Or, more simply: $L(x) \approx \sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625} .$

Answer: $L(x) \approx \sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625} \approx 0.373082337744149$

A quadratic approximation is given by: $Q(x) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{1}{2} f^{\prime \prime}\left(x_{0}\right)\left(x-x_{0}\right)^{2} .$

We are given that $x_{0}=0 .$

Firstly, find the value of the function at the given point:

$y_{0}=f\left(x_{0}\right)=\sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625}$

Secondly, find the derivative of the function, evaluated at the point: $f^{\prime}(0) .$

Find the derivative: $f^{\prime}(x)=0$

Next, evaluate the derivative at the given point.

$f^{\prime}(0)=0$

Now, find the second derivative of the function evaluated at the point: $f^{\prime \prime}(0) .$

Find the second derivative: $f^{\prime \prime}(x)=0$

Next, evaluate the second derivative at the given point.

$f^{\prime \prime}(0)=0 .$

Plugging the found values, we get that:

$Q(x) \approx \sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625}+0(x-(0))+\frac{1}{2}(0)(x-(0))^{2}\\ Simplify: Q(x) \approx \sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625} .\\ Answer: Q(x) \approx \sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625}$

(b) Actual Value of function is: $\sin \left(\frac{\pi}{25}\right) \tan \left(\frac{13 \pi}{50}\right)+\frac{3744}{15625}$

So, we can say that actual value of function is same as linear and quadratic approximation's value because no variable is there.