The base is always \(4\) times the perpendicular height.
Find the rate of change of the area of the triangle with respect to its vertical height when the area of the triangle is \(98\text{ cm}^2\).
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
Start by writing the area in terms of height, use the area \(98\) to find the height, then differentiate and evaluate the rate when the height is \(7\).
Step 1
Write the base in terms of height
If the height is \(h\), what is the base?
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Step 2
Form the area expression
Using \(A=\frac{1}{2}bh\), which expression gives the area in terms of \(h\)?
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Step 3
Find the height when the area is 98
Substitute \(A=98\) into the area formula. What is the height?
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Step 4
Differentiate the area function
What is \(\frac{dA}{dh}\) if \(A=2h^2\)?
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Step 5
Evaluate the rate at \(h=7\)
Substitute \(h=7\) into the derivative. What is the rate of change?
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Step 6
Final answer
Now write the method and answer clearly.
\[
b=4h
\]
\[
A=\frac{1}{2}bh
\]
\[
A=\frac{1}{2}(4h)(h)=2h^2
\]
\[
98=2h^2
\]
\[
h^2=49
\]
\[
h=7
\]
\[
\frac{dA}{dh}=4h
\]
\[
\frac{dA}{dh}\Big|_{h=7}=4(7)=28
\]
\[
\text{The rate of change of area with respect to height is }28\text{ cm}^2\text{ per cm.}
\]
Great work. When the area is \(98\text{ cm}^2\), the height is \(7\) cm and the rate of change is \(28\text{ cm}^2\) per cm.