Find the gradient of the tangent to the curve [tex]y=\frac{x \text { sinx }}{\cos x+x}[/tex] at the point [tex]x=\frac{\pi}{3}[/tex]
Answer (1)
Find the derivative of y = c o s x + x x s i n x using the quotient rule: d x d y = ( c o s x + x ) 2 ( c o s x + x ) ( s i n x + x c o s x ) − ( x s i n x ) ( 1 − s i n x ) .
Substitute x = 3 π into the derivative expression.
Evaluate the expression using sin ( 3 π ) = 2 3 and cos ( 3 π ) = 2 1 .
The gradient of the tangent at x = 3 π is approximately 0.8474 .
Explanation
Problem Setup We are asked to find the gradient of the tangent to the curve y = c o s x + x x s i n x at the point x = 3 π . The gradient of the tangent is given by the derivative of y with respect to x , i.e., d x d y .
Applying the Quotient Rule To find the derivative d x d y , we will use the quotient rule. The quotient rule states that if y = v u , then d x d y = v 2 v d x d u − u d x d v . In this case, u = x sin x and v = cos x + x .
Finding Derivatives of u and v Now we need to find d x d u and d x d v . For u = x sin x , we use the product rule: d x d u = sin x + x cos x . For v = cos x + x , we have d x d v = − sin x + 1 .
Substituting into Quotient Rule Substituting these into the quotient rule formula, we get d x d y = ( cos x + x ) 2 ( cos x + x ) ( sin x + x cos x ) − ( x sin x ) ( 1 − sin x )
Evaluating at x = pi/3 Now we need to evaluate d x d y at x = 3 π . Plugging in x = 3 π , we have d x d y x = 3 π = ( cos ( 3 π ) + 3 π ) 2 ( cos ( 3 π ) + 3 π ) ( sin ( 3 π ) + 3 π cos ( 3 π )) − ( 3 π sin ( 3 π )) ( 1 − sin ( 3 π ))
Substituting Trigonometric Values We know that sin ( 3 π ) = 2 3 and cos ( 3 π ) = 2 1 . Substituting these values, we get d x d y x = 3 π = ( 2 1 + 3 π ) 2 ( 2 1 + 3 π ) ( 2 3 + 3 π ⋅ 2 1 ) − ( 3 π ⋅ 2 3 ) ( 1 − 2 3 )
Calculating the Result After calculating the above expression, we find that the gradient of the tangent at x = 3 π is approximately 0.8474.
Final Answer Therefore, the gradient of the tangent to the curve at x = 3 π is approximately 0.8474 .
Examples
Imagine you are designing a rollercoaster. The gradient of the tangent at a specific point on the track is crucial for calculating the instantaneous speed and acceleration of the cart. By finding the derivative of the track's equation, you can ensure the rollercoaster's safety and thrill factor at different points along the ride. This problem demonstrates how calculus is essential in engineering design to predict and control dynamic systems.
