Find the gradient of the tangent to the curve [tex]$w=\frac{v^3-7 v+2}{(v+2)(v^2-v+3)}$[/tex] at the point [tex]$v=\frac{2}{3}$[/tex]
Answer (1)
Find the derivative d v d w of the given function w = ( v + 2 ) ( v 2 − v + 3 ) v 3 − 7 v + 2 using the quotient rule.
Evaluate the derivative at the point v = 3 2 .
Simplify the expression to find the gradient of the tangent.
The gradient of the tangent to the curve at v = 3 2 is − 5000 3033 .
Explanation
Problem Setup We are given the curve w = ( v + 2 ) ( v 2 − v + 3 ) v 3 − 7 v + 2 and we want to find the gradient of the tangent to this curve at the point v = 3 2 . The gradient of the tangent is given by the derivative d v d w evaluated at v = 3 2 .
Finding the Derivative First, we need to find the derivative of w with respect to v . We have
w = ( v + 2 ) ( v 2 − v + 3 ) v 3 − 7 v + 2 = v 3 + v 2 + v + 6 v 3 − 7 v + 2
Using the quotient rule, we have
d v d w = ( v 3 + v 2 + v + 6 ) 2 ( 3 v 2 − 7 ) ( v 3 + v 2 + v + 6 ) − ( v 3 − 7 v + 2 ) ( 3 v 2 + 2 v + 1 )
Evaluating the Derivative at v=2/3 Now we need to evaluate the derivative at v = 3 2 . Plugging in v = 3 2 into the expression for d v d w , we get
d v d w ∣ v = 3 2 = (( 3 2 ) 3 + ( 3 2 ) 2 + ( 3 2 ) + 6 ) 2 ( 3 ( 3 2 ) 2 − 7 ) (( 3 2 ) 3 + ( 3 2 ) 2 + ( 3 2 ) + 6 ) − (( 3 2 ) 3 − 7 ( 3 2 ) + 2 ) ( 3 ( 3 2 ) 2 + 2 ( 3 2 ) + 1 )
d v d w ∣ v = 3 2 = (( 27 8 ) + ( 9 4 ) + ( 3 2 ) + 6 ) 2 ( 3 ( 9 4 ) − 7 ) (( 27 8 ) + ( 9 4 ) + ( 3 2 ) + 6 ) − (( 27 8 ) − 7 ( 3 2 ) + 2 ) ( 3 ( 9 4 ) + 2 ( 3 2 ) + 1 )
d v d w ∣ v = 3 2 = (( 27 8 ) + ( 27 12 ) + ( 27 18 ) + 27 162 ) 2 ( 3 4 − 7 ) ( 27 8 + 27 12 + 27 18 + 27 162 ) − ( 27 8 − 3 14 + 2 ) ( 3 4 + 3 4 + 1 )
d v d w ∣ v = 3 2 = ( 27 200 ) 2 ( 3 4 − 21 ) ( 27 200 ) − ( 27 8 − 126 + 54 ) ( 3 8 + 3 )
d v d w ∣ v = 3 2 = ( 27 200 ) 2 ( 3 − 17 ) ( 27 200 ) − ( 27 − 64 ) ( 3 11 )
d v d w ∣ v = 3 2 = 729 40000 81 − 3400 + 81 704 = 729 40000 81 − 2696 = 81 − 2696 × 40000 729 = 40000 − 2696 × 9 = 40000 − 24264 = 5000 − 3033
So, the gradient of the tangent at v = 3 2 is − 5000 3033 .
Final Answer The gradient of the tangent to the curve at v = 3 2 is − 5000 3033 .
Examples
In physics, understanding the gradient of a tangent to a curve is essential in analyzing motion. For example, if the curve represents the position of an object over time, the gradient of the tangent at a specific time gives the object's instantaneous velocity at that moment. This concept is crucial in fields like robotics, where precise control of movement requires accurate velocity calculations. By determining the gradient, engineers can fine-tune the robot's motion to perform tasks efficiently and safely.
