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Question 8
The curve shown in Figure 3 has parametric equations $x = t - 4 \, ext{sin} \, t, \ y = 1 - 2 \, \text{cos} \, t, \quad \frac{2 \pi}{3} \leq t \leq \frac{2 \pi}{3}... show full transcript
Step 1
Answer
To find the exact value of , we start by substituting into the equation for :
Rearranging gives:
So, ( \text{cos} , t = 0 ), which occurs at:
Given the constraint , the possible value of is:
Now substituting back into the equation for :
As :
Therefore, the exact value of is:
Step 2
Answer
To find the gradient of the curve at point where , we first calculate ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ):
Next, we evaluate ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) at :
For :
For :
Thus, the gradient () of the curve is given by:
Step 3
Answer
To find the value of where the gradient is :
Given the equation for the gradient:
Cross-multiplying results in:
This simplifies to:
Dividing through by 4 gives:
Using the identity and substituting :
Expanding and simplifying:
This leads to:
Factoring out :
So or . The solution gives:
Next, consider the value of :
Consolidating the numerical value we find:
(4 decimal places) Thus, the value of is approximately .
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