fbpx
calculus online - Free exercises and solutions to help you succeed!

Multivariable Linear Approximation – A multiplication of sin and tan functions in 2 variables – Exercise 4211

Exercise

Find an approximate value of

\sin 32^{\circ} \tan 40^{\circ}

Final Answer


\frac{85+\sqrt{3}\pi}{180}

Solution

We will use the 2-variable linear approximation formula

f(x,y)\approx f(x_0,y_0)+f'_x(x_0,y_0)\cdot(x-x_0)+f'_y(x_0,y_0)\cdot(y-y_0)

Therefore, we need to define the following

x, y, x_0, y_0, f(x)

And place them in the formula. x and y will be the numbers that appear in the question, and the points

x_0,y_0

will be the closest values to x and y respectively, which are easy for calculations.

In our exercise, we will define

x=32^{\circ}, y=40^{\circ}

because these values appear in the question. And we set

x_0=30^{\circ}=\frac{\pi}{6}

y_0=45^{\circ}=\frac{\pi}{4}

because they are the closest values to x and y respectively that are easy for calculations.

After we have defined x and y, it is easy to find the function. Just put x instead of the number we set to be x and put y instead of the number we set to be y. In this way, we get the function

f(x,y)=\sin x\cdot\tan y

In the formula above we see the function partial derivatives. Hence, we calculate them.

f'_x(x,y)=\cos x\cdot \tan y

f'_y(x,y)=\sin x\cdot\frac{1}{\cos^2 y}

We put all the data in the formula and get

f(\frac{\pi}{6},\frac{\pi}{4})\approx f(\frac{\pi}{6},\frac{\pi}{4})+f'_x(\frac{\pi}{6},\frac{\pi}{4})\cdot(32^{\circ}-30^{\circ})+f'_y(\frac{\pi}{6},\frac{\pi}{4})\cdot(40^{\circ}-45^{\circ})=

=\sin \frac{\pi}{6}\cdot\tan\frac{\pi}{4}+\cos\frac{\pi}{6}\cdot\tan\frac{\pi}{4}\cdot 2^{\circ}+\sin\frac{\pi}{6}\cdot\frac{1}{\cos^2(\frac{\pi}{4})}\cdot(-5^{\circ})=

=\frac{1}{2}\cdot 1+\frac{\sqrt{3}}{2}\cdot 1\cdot\frac{\pi}{90}+\frac{1}{2}\cdot\frac{1}{{(\frac{\sqrt{2}}{2})}^2}\cdot(-\frac{\pi}{36})=

=\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\pi}{90}+\frac{1}{2}\cdot 2\cdot(-\frac{\pi}{36})=

=\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\pi}{90}-\frac{\pi}{36}=

=\frac{1}{2}+\frac{\sqrt{3}\pi}{180}-\frac{\pi}{36}=

=\frac{90+\sqrt{3}\pi-5}{180}=

=\frac{85+\sqrt{3}\pi}{180}

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply